Work Questions in English

(A) When a force $F$ acts on an object to displace it through a distance $S$ in its direction,the work done $W$ on the body by the force is given by:
Work done = Force $\times$ Displacement
$W = F \times S$
Given:
Force $(F)$ = $7\, N$
Displacement $(S)$ = $8\, m$
Therefore,work done $(W)$ = $7\, N \times 8\, m$
$W = 56\, N\cdot m$
Since $1\, N\cdot m = 1\, J$,the work done is $56\, J$.

(C) Work is said to be done when the following two conditions are satisfied:
$(i)$ $A$ force acts on the body.
$(ii)$ The body undergoes displacement due to the applied force,and the displacement occurs in the direction of the force.

(N/A) When a force $F$ displaces a body through a distance $s$ in the direction of the applied force,the work done $W$ on the body is defined as the product of the force and the displacement.
Work done = Force $\times$ Displacement
$W = F \times s$

(N/A) $1\, J$ is defined as the amount of work done when a force of $1\, N$ displaces an object through a distance of $1\, m$ in the direction of the applied force.
Mathematically,$W = F \times s$,where $W = 1\, J$,$F = 1\, N$,and $s = 1\, m$.

(A) Work done by the bullocks is calculated using the formula:
Work done = Force $\times$ Displacement
$W = F \times d$
Given:
Applied force,$F = 140 \, N$
Displacement,$d = 15 \, m$
Substituting the values into the formula:
$W = 140 \, N \times 15 \, m = 2100 \, J$
Therefore,the total work done in ploughing the length of the field is $2100 \, J$.

(N/A) $1$. Suma is swimming in a pond: Suma is doing work as she is able to move herself by applying force with the movement of her arms and legs in the water.
$2$. $A$ donkey is carrying a load on its back: The donkey is not doing any work (in the sense of physics) as the weight he is carrying (the direction of force) and the displacement are perpendicular to each other $(W = Fs \cos 90^{\circ} = 0)$.
$3$. $A$ windmill is lifting water from a well: The windmill is lifting water from a well and doing work against gravity.
$4$. $A$ green plant is carrying out photosynthesis: No force and displacement are present here,so the work done is zero.
$5$. An engine is pulling a train: During the pulling of a train,the engine does work against the friction present between the railway track and the wheels.
$6$. Food grains are getting dried in the sun: During the drying of grains,there is no force applied,nor is there any displacement. So,no work is done.
$7$. $A$ sailboat is moving due to wind energy: Work is done by the wind as it moves the sailboat in the direction of the force (force of blowing air).

(A) The work done by a force is given by the formula $W = F \cdot s \cdot \cos(\theta)$, where $F$ is the force, $s$ is the displacement, and $\theta$ is the angle between the force and displacement vectors.
In this case, the force of gravity acts vertically downwards $(F = mg)$.
The displacement $s$ is the straight-line distance between the initial and final points.
Since the initial and final points lie on the same horizontal line, the vertical displacement is $0$.
Because the gravitational force acts only in the vertical direction, the work done by gravity is $W = F \times (\text{vertical displacement}) = mg \times 0 = 0 \ J$.

(N/A) The work done by the gravitational force acting on the body is zero.
This is because the gravitational force acts vertically downward,while the displacement of the object is horizontal.
Since the angle $\theta$ between the force and the displacement is $90^\circ$,the work done $W$ is calculated as:
$W = F \cdot s \cdot \cos(90^\circ)$
Since $\cos(90^\circ) = 0$,the work done $W = 0$.

(N/A) In physics,work is defined as the product of force and displacement $(W = F \times s)$. Since the rock does not move,the displacement $(s)$ is $0$,meaning no mechanical work is done on the rock.
However,energy is still being expended by your body. When you push the rock,your muscles contract and perform internal work. This energy is primarily dissipated as heat into the environment and the contact surface between your hands and the rock. $A$ negligible amount may be stored as elastic potential energy due to the slight deformation of the rock's surface.

(C) When a satellite moves around the Earth in a circular orbit,the gravitational force acts towards the center of the Earth (centripetal force).
The displacement of the satellite at any instant is along the tangent to the circular path.
Since the radius (direction of force) is always perpendicular to the tangent (direction of displacement),the angle $\theta$ between the force and displacement is $90^{\circ}$.
The formula for work done is $W = F \cdot s \cdot \cos(\theta)$.
Substituting $\theta = 90^{\circ}$,we get $W = F \cdot s \cdot \cos(90^{\circ}) = F \cdot s \cdot 0 = 0$.
Therefore,the work done by the force of gravity on a satellite moving around the Earth is zero.

(N/A) In scientific terms,the person has done no work.
Work is defined as the product of the force applied and the displacement in the direction of the force $(W = F \times s)$.
Since the bundle of hay is held stationary over the head,the displacement $(s)$ is $0$.
Therefore,$W = F \times 0 = 0$.
The person feels tired because his muscular energy is being consumed and converted into thermal energy to maintain the posture,but this does not constitute mechanical work.

(N/A) The work done by a force is given by the formula $W = F \cdot s \cdot \cos(\theta)$,where $\theta$ is the angle between the force and the displacement.
$1$. In the first case,the force $F$ is acting vertically upwards while the displacement is horizontal (west to east). The angle between them is $90^{\circ}$. Since $\cos(90^{\circ}) = 0$,the work done is zero.
$2$. In the second case,the force $F$ is acting in the same direction as the displacement (west to east). The angle between them is $0^{\circ}$. Since $\cos(0^{\circ}) = 1$,the work done is positive.
$3$. In the third case,the force $F$ is acting in the opposite direction to the displacement (east to west). The angle between them is $180^{\circ}$. Since $\cos(180^{\circ}) = -1$,the work done is negative.

(C) Work done is defined by the formula $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force,$s$ is the displacement,and $\theta$ is the angle between them.
When force and displacement are in mutually opposite directions,the angle $\theta$ is $180^o$.
Since $\cos(180^o) = -1$,the work done becomes $W = -F \cdot s$,which is negative work.

(A) Work done is defined as the product of force and displacement in the direction of the force,given by $W = F \cdot s \cdot \cos(\theta)$.
Here,the gravitational force acts vertically downwards on the bag.
The displacement of the girl is along the horizontal levelled road.
Since the angle between the gravitational force (downward) and the displacement (horizontal) is $90^{\circ}$,the work done against gravity is $W = F \cdot s \cdot \cos(90^{\circ})$.
Since $\cos(90^{\circ}) = 0$,the work done against the gravitational force is $0 \, J$.

(C) The formula for work done is $W = F \cdot s \cdot \cos(\theta)$,where $W$ is work,$F$ is the force applied,$s$ is the displacement,and $\theta$ is the angle between the force and the displacement vector.
From this formula,it is clear that work done depends on the magnitude of the force,the magnitude of the displacement,and the angle between them.
Work done does not depend on the initial velocity of the object.

(POSITIVE) The work done is positive. This is because the man applies a force in the upward direction to lift the bucket,and the displacement of the bucket also occurs in the same upward direction. Since the angle between the force vector and the displacement vector is $0^{\circ}$,the work done $(W = F \cdot s \cdot \cos(0^{\circ}))$ is positive.

(B) The work done by the gravitational force is negative.
This is because the displacement of the bucket is in the upward direction,while the gravitational force acts in the downward direction.
Since the angle between the force and the displacement is $180^{\circ}$,the work done $W = F \cdot s \cdot \cos(180^{\circ}) = -F \cdot s$,which is negative.

(NEGATIVE) The work done by friction is negative.
Reasoning: Work done $(W)$ is defined by the formula $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force,$s$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
When a body slides down an inclined plane,the direction of motion (displacement) is downwards along the plane.
Frictional force always acts in the direction opposite to the motion of the body.
Therefore,the angle $\theta$ between the frictional force and the displacement is $180^{\circ}$.
Since $\cos(180^{\circ}) = -1$,the work done $W = F \cdot s \cdot (-1) = -F \cdot s$,which is negative.

(A) When a body moves on a rough horizontal plane with uniform velocity,the net force acting on it is zero.
This implies that the applied force is equal in magnitude and opposite in direction to the frictional force.
The work done by the applied force is positive because the applied force acts in the direction of the displacement of the body.
Conversely,the work done by the frictional force is negative because it acts opposite to the direction of displacement.

(NEGATIVE) The work done by the resistive force of air on a vibrating pendulum is negative.
This is because the resistive force of air (air resistance) always acts in a direction opposite to the direction of the motion of the pendulum.
Since the angle between the force and the displacement is $180^{\circ}$,the work done $(W = F \cdot s \cdot \cos(180^{\circ}))$ results in a negative value.

(N/A) Negative work occurs when the force applied is in the opposite direction to the displacement of the object. An example is the work done by the force of friction on a moving object. As the object moves in one direction,the frictional force acts in the opposite direction,resulting in an angle of $180^{\circ}$ between the force and displacement,which makes the work done negative $(W = F \cdot s \cdot \cos(180^{\circ}) = -Fs)$.

(B) The work done on a body is defined by the formula $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force applied,$s$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
Work done depends on the force applied and the displacement caused by that force.
It does not depend on the velocity of the body,as velocity is the rate of change of displacement,not a factor in the definition of mechanical work.

(C) The work done by a force is given by the formula $W = F \cdot s \cdot \cos(\theta)$,where $\theta$ is the angle between the force and the displacement.
In this case,the weight of the body acts vertically downwards,while the displacement of the body is horizontal (along the surface).
Since the angle between the vertical weight and the horizontal displacement is $90^\circ$,we have $\cos(90^\circ) = 0$.
Therefore,the work done by the weight is $W = F \cdot s \cdot 0 = 0$.

(C) Work done $(W)$ is defined as the product of force $(F)$ and displacement $(s)$ in the direction of the force,given by the formula $W = F \times s \times \cos(\theta)$.
Since the person fails to move the heavy mass,the displacement $(s)$ is $0$.
Therefore,the work done is $W = F \times 0 = 0 \text{ Joules}$.
Thus,the work done by the person is zero.

(B) The work done $(W)$ in raising an object vertically is given by the formula $W = mgh$,where $m$ is the mass of the object,$g$ is the acceleration due to gravity,and $h$ is the height to which the object is raised.
Since the work done depends only on the force applied (which is equal to the weight of the object,$mg$) and the displacement $(h)$,it is independent of the time taken or the speed at which the object is raised.
Therefore,the work done does not depend on how fast the object is raised.

(N/A) Work is defined as the product of force and displacement in the direction of the force. When a porter moves vertically upstairs,he applies an upward force to balance the weight of the load (acting downwards). Since the displacement of the porter is also in the upward direction,the angle between the force applied by the porter and the displacement is $0^{\circ}$. Therefore,the work done by the porter is positive and non-zero $(W = F \cdot s \cdot \cos(0^{\circ}) = F \cdot s)$.

(C) In a simple pendulum,the tension force $T$ in the string always acts towards the point of suspension.
As the pendulum bob moves along its circular arc,the displacement vector $ds$ is always tangent to the path.
The tension force $T$ is always directed towards the center of the circle,which is perpendicular to the displacement vector $ds$ at every point.
Since the work done $W$ is defined by the dot product $W = \int T \cdot ds = \int T \cos(90^{\circ}) ds$,and $\cos(90^{\circ}) = 0$,the work done by the tension is $0$.

(N/A) In a tug-of-war,the winning team exerts a force on the losing team,causing them to move in the direction of the force.
Since the force applied by the winning team and the displacement of the losing team are in the same direction,the winning team performs positive work.
The magnitude of the work done is equal to the product of the resultant force applied by the winning team and the displacement undergone by the losing team.

(A) In physics,work is defined as the product of the force applied and the displacement caused in the direction of the force. The formula for work is $W = F \times s \times \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
Since the rock did not move,the displacement $(s)$ is $0 \, m$.
Therefore,$W = F \times 0 = 0 \, J$.
Thus,the work done by Seema is $0 \, J$.

(C) The formula for work done is $W = F \cdot s \cdot \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
If the displacement $(s)$ of the object is $0$,then $W = F \cdot 0 \cdot \cos(\theta) = 0$.
Therefore,the work done on the object is $0$.

(A) Work done is defined as the product of force and displacement in the direction of force,given by the formula $W = F \times S \times \cos(\theta)$.
In this scenario,the weight lifter is holding the weight stationary on his shoulders.
Since there is no change in the position of the weight,the displacement $(S)$ is $0$.
Therefore,$W = F \times 0 = 0$.
Thus,the work done by the weight lifter is $0 \,J$.

(N/A) The formula for calculating work done is $W = F \times S \times \cos(\theta)$,where $W$ is work,$F$ is the magnitude of the force applied,$S$ is the displacement,and $\theta$ is the angle between the force and displacement vectors. In the simplest case where force is applied in the direction of displacement,the formula is $W = F \times S$.
The $SI$ unit of work is the joule $(J)$,where $1 \ J = 1 \ N \cdot m$.

(C) The work done $(W)$ is defined by the formula $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force,$s$ is the displacement,and $\theta$ is the angle between the force and the displacement.
In the case of the earth moving around the sun,the gravitational force acts towards the center of the sun (centripetal force),while the displacement of the earth is along the tangent to its circular orbit.
Since the force and the displacement are perpendicular to each other,the angle $\theta = 90^\circ$.
Therefore,$W = F \cdot s \cdot \cos(90^\circ) = F \cdot s \cdot 0 = 0$.
Thus,the work done by the earth in moving around the sun is zero.

(A) Work is defined as the product of the force applied and the displacement in the direction of the force.
Mathematically,$W = F \times s \times \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
In this scenario,the person is standing still,which means the displacement $s = 0 \, m$.
Since the displacement is zero,the work done $W = F \times 0 = 0 \, J$.
Therefore,no work is done by the person.

(N/A) Work is defined as the product of the magnitude of the force applied on an object and the displacement of the object in the direction of the force.
The expression for work done $(W)$ is given by:
$W = F \times s$
Where $F$ is the force applied and $s$ is the displacement.
The $SI$ unit of work done is the joule $(J)$,where $1 \ J = 1 \ N \cdot m$.

(A) The work done in lifting an object against gravity is given by the formula: $W = mgh$.
Here,mass $m = 40 \text{ kg}$,height $h = 0.5 \text{ m}$,and acceleration due to gravity $g = 9.8 \text{ m s}^{-2}$.
Substituting these values into the formula:
$W = 40 \times 9.8 \times 0.5$
$W = 40 \times 4.9$
$W = 196 \text{ Joules}$.
Therefore,the total work done is $196 \text{ J}$.

(N/A) The work done $(W)$ by a force is defined by the formula $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force,$s$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
Therefore,the work done depends on the following three factors:
$(i)$ The magnitude of the force $(F)$ applied to the object.
$(ii)$ The magnitude of the displacement $(s)$ produced in the object.
$(iii)$ The angle $(\theta)$ between the direction of the applied force and the direction of the displacement.

(C) In order to balance the load on his head,the coolie applies a force on it in the upward direction,equal to its weight.
His displacement is along the horizontal direction.
Thus,the angle between the force $F$ and the displacement $S$ is $90^{\circ}$.
Therefore,the work done $W = F S \cos \theta = F S \cos 90^{\circ} = 0$.

(N/A) The work done $(W)$ in lifting a box of mass $(m)$ through a vertical height $(h)$ against the gravitational force $(F = mg)$ is given by the formula $W = F \times h = mgh$.
In this expression,the work done depends only on the mass of the object $(m)$,the acceleration due to gravity $(g)$,and the vertical displacement $(h)$.
Since the time taken or the speed at which the box is lifted does not appear in this formula,the work done is independent of the rate at which the box is lifted.

(N/A) Yes,it is possible when the force acting on the body is perpendicular to the direction of its motion.
Work done $(W)$ is defined as $W = Fs \cos \theta$,where $\theta$ is the angle between the force $(F)$ and the displacement $(s)$.
If $\theta = 90^{\circ}$,then $\cos 90^{\circ} = 0$,resulting in $W = 0$.
An example of this is the moon revolving around the earth in a circular orbit. The earth exerts a gravitational centripetal force on the moon,which acts towards the center of the orbit,while the moon's velocity is tangential to the orbit. Since the force is always perpendicular to the displacement,the work done by the earth on the moon is zero,even though the moon is in accelerated motion due to the change in the direction of its velocity.

(0) When a body moves along a circular path,the centripetal force acting on it is always directed towards the center of the circle (along the radius).
The displacement of the body at any instant is along the tangent to the circular path at that point.
Since the radius and the tangent at any point on a circle are perpendicular to each other,the angle $\theta$ between the force $(F)$ and the displacement $(S)$ is $90^{\circ}$.
The formula for work done is $W = F S \cos \theta$.
Substituting $\theta = 90^{\circ}$,we get $W = F S \cos 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the work done $W = 0$.

(NO) Work is defined as the product of force and displacement in the direction of the force $(W = F \cdot d \cdot \cos \theta)$.
In this scenario,the man is applying force to the oars to move the boat relative to the water,so he is doing work against the stream.
However,with respect to the shore,the boat's displacement $(d)$ is $0$ because the man is at rest relative to the shore.
Since $W = F \cdot 0 = 0$,the man does no work with respect to the shore.

(N/A) Case $I$: The force $F$ is acting in the vertically upward direction,while the displacement of the object is in the horizontal direction (from west to east). Since the force and displacement are perpendicular to each other (angle $\theta = 90^\circ$),the work done is $W = F \cdot s \cdot \cos(90^\circ) = 0$.
Case $II$: Here,the force $F$ is applied in the same direction as the displacement. Since the angle between force and displacement is $0^\circ$,the work done is positive $(+ve)$,as $W = F \cdot s \cdot \cos(0^\circ) = F \cdot s$.

(N/A) Work done is calculated by the formula $W = F \times s$. Given $F = 1\, N$ and $s = 1\, m$,the work done is $1\, N \times 1\, m = 1\, J$.
$(b)$ Yes,it is possible for the work done to be zero even if a force is acting on a body. This occurs when the force is applied at an angle of $90^{\circ}$ to the direction of displacement. Since $W = Fs \cos(\theta)$,if $\theta = 90^{\circ}$,then $\cos(90^{\circ}) = 0$,making the work done zero.
Example: When a satellite moves in a circular orbit around the Earth,the gravitational force acts towards the center of the Earth (perpendicular to the motion),while the displacement is along the tangent to the orbit. Thus,the work done by gravity on the satellite is zero.

(N/A) Work done is given by the formula $W = F S \cos \theta$,where $\theta$ is the angle between the force vector $F$ and the displacement vector $S$.
Work done is said to be negative when the angle $\theta$ between the force and the displacement is obtuse,i.e.,$90^{\circ} < \theta \leq 180^{\circ}$,because $\cos \theta$ is negative in this range.
$A$ common situation is a body falling vertically downwards through the air. In this case,the force of gravity acts in the direction of motion (downwards),so the work done by gravity is positive. Simultaneously,the air resistance (frictional force) acts in the opposite direction of motion (upwards),so the work done by air resistance is negative.

(N/A) For work to be done on an object,the following two conditions must be satisfied:
$(i)$ $A$ force must be applied to the object.
$(ii)$ The object must undergo a displacement in the direction of the force (or have a component of displacement in the direction of the force,i.e.,the angle $\theta$ between force and displacement must be such that $\cos \theta \neq 0$).

(A) The mass of the box is $m = 10 \, kg$.
The gravitational force (weight) acting on the box is $F = mg = 10 \, kg \times 10 \, m s^{-2} = 100 \, N$,which acts vertically downwards.
The displacement $S = 2 \, m$ is along the horizontal surface.
The angle $\theta$ between the gravitational force (downward) and the displacement (horizontal) is $90^{\circ}$.
The work done $W$ is given by the formula $W = FS \cos \theta$.
Substituting the values: $W = 100 \times 2 \times \cos 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the work done $W = 100 \times 2 \times 0 = 0 \, J$.
Therefore,the work done by the gravitational force is $0 \, J$ because the force of gravity is perpendicular to the direction of displacement.

(II, III) In physics,work is said to be done when a force applied on an object causes a displacement in the direction of the force. The formula is $W = F \times s \times \cos(\theta)$.
$(i)$ No work is done because the displacement $(s)$ is $0$,even though a force is applied.
$(ii)$ Work is done because the bullock applies force to the cart,resulting in a displacement of $1 \, km$.
$(iii)$ Work is done because the girl applies force to the trolley,resulting in a displacement of $2 \, m$.
$(iv)$ No work is done because the person is stationary,meaning the displacement $(s)$ is $0$.

(A) Work done by a force is measured as the product of the magnitude of the force applied and the displacement in the direction of the force $(W = F \times S)$.
Given:
Mass of the luggage $(m)$ = $20 \, kg$
Displacement $(h)$ = $1.7 \, m$
Acceleration due to gravity $(g)$ = $10 \, m s^{-2}$
The force applied by the porter to lift the luggage is equal to the weight of the luggage,which is $F = m \times g$.
Therefore,the work done $(W)$ = $m \times g \times h$.
$W = 20 \, kg \times 10 \, m s^{-2} \times 1.7 \, m$.
$W = 340 \, J$.
Thus,the work done by the porter on the luggage is $340 \, J$.

(N/A) Work is defined as the product of the force applied to an object and the displacement produced in the direction of the force. Mathematically,$W = F \times s \times \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
Work is measured as the product of the magnitude of force and the displacement in the direction of the force.
Work done by a force is negative when the force acts in the direction opposite to the direction of displacement,i.e.,the angle between force and displacement is $180^{\circ}$ (since $\cos(180^{\circ}) = -1$).