Work Questions in English

(N/A) Ram is not justified in making the statement that he has done a lot of work.
In physics,the work done on an object is defined as the product of the force applied and the displacement produced in the direction of the force $(W = F \times s)$.
For work to be done,two conditions must be satisfied:
$1$. $A$ force must be applied to the object.
$2$. The object must undergo displacement.
Since the wall does not move,the displacement $(s)$ is $0$. Therefore,the work done on the wall is $W = F \times 0 = 0$.
Although Ram feels exhausted,this is due to internal physiological processes where his muscles contract and expand,consuming chemical energy,but no external mechanical work is performed on the wall.

(N/A) The work done by the force applied by the boy is positive because the displacement of the ball is in the same direction as the applied force.
$(b)$ The work done by the gravitational force is negative because the gravitational force acts downwards while the displacement of the ball is upwards,meaning the force and displacement are in opposite directions.

(N/A) The force applied to lift the box is equal to its weight,$F = mg = 250 \,kg \times 10 \,m/s^2 = 2500 \,N$. The displacement is $S = 1 \,m$. The work done is $W = F \times S = 2500 \,N \times 1 \,m = 2500 \,J$.
$(b)$ When holding the box,there is no displacement $(S = 0)$. Therefore,the work done is $W = F \times 0 = 0 \,J$.
$(c)$ They get tired because of the continuous muscular effort required to counteract the gravitational force acting on the box,even though no mechanical work is performed.

(N/A) The work done by the man is equal to the change in the potential energy of the system (man + weight).
The total mass $m$ is the sum of the man's mass and the weight he carries:
$m = 70 \,kg + 10 \,kg = 80 \,kg$
The height $h$ is $100 \,m$.
Taking the acceleration due to gravity $g = 9.8 \,m/s^2$,the work done $W$ is calculated as:
$W = m \times g \times h$
$W = 80 \,kg \times 9.8 \,m/s^2 \times 100 \,m$
$W = 78400 \,J$ (or $7.84 \times 10^4 \,J$).

(B) The work done in moving a body through a certain distance is given by the formula:
$W = F \times S \times \cos \theta$
Assuming the force is applied in the direction of displacement,$\theta = 0^\circ$,so $\cos 0^\circ = 1$.
Given:
Force $(F)$ = $100 \, N$
Displacement $(S)$ = $10 \, m$
Mass $(m)$ = $50 \, kg$ (Note: Mass is not required for this calculation as force is already provided).
Calculation:
$W = F \times S = 100 \, N \times 10 \, m = 1000 \, J$.

(C) Work done is defined as $W = F S \cos \theta$,where $F$ is the force,$S$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
$1$. For the horizontal motion: The force applied is vertical ($10 \,N$ upwards) and the displacement is horizontal $(5 \,m)$. The angle $\theta = 90^{\circ}$.
$W_{1} = F S \cos 90^{\circ} = 10 \times 5 \times 0 = 0 \,J$.
$2$. For the vertical motion: The force applied is vertical ($10 \,N$ upwards) and the displacement is vertical ($10 \,m$ upwards). The angle $\theta = 0^{\circ}$.
$W_{2} = F S \cos 0^{\circ} = 10 \times 10 \times 1 = 100 \,J$.
Total work done = $W_{1} + W_{2} = 0 + 100 = 100 \,J$.

(A) To lift the car,the crane must perform work against the force of gravity.
The force required is equal to the weight of the car: $F = mg = 500 \, kg \times 9.8 \, m/s^2 = 4900 \, N$.
The displacement of the car is $S = 4 \, m$.
Work done is calculated as: $W = F \times S = 4900 \, N \times 4 \, m = 19600 \, J$.

(A) The work done by the man is against the gravitational force acting on the weight he is lifting.
Here,the mass of the object being lifted is $m = 30 \, kg$.
The height to which it is lifted is $h = 10 \, m$.
The acceleration due to gravity is $g = 9.8 \, m s^{-2}$.
The work done $W$ is given by the formula $W = mgh$.
Substituting the values: $W = 30 \times 9.8 \times 10$.
$W = 30 \times 98 = 2940 \, J$.
Note: The weight of the man $(50 \, kg)$ is irrelevant to the work done on the object.

(D) $(i)$ The work done against the force of gravity is equal to the change in gravitational potential energy,which is given by the formula $W = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the height.
$(ii)$ Given: mass $(m) = 45 \, kg$,height $(h) = 10 \, m$,and acceleration due to gravity $(g) \approx 10 \, m/s^2$.
Work done $(W) = m \times g \times h$
$W = 45 \, kg \times 10 \, m/s^2 \times 10 \, m$
$W = 4500 \, J$.

(B) Work is defined as the product of force and displacement in the direction of the force $(W = F \times s)$.
In the first case,the boy pushes the wall,but since the wall does not move,the displacement $(s)$ is $0$. Therefore,the work done on the wall is $W = F \times 0 = 0$.
In the second case,the boy pushes the chair,and it moves from the front to the back of the class,meaning there is a non-zero displacement $(s > 0)$.
Since work is done on the chair and zero work is done on the wall,more work is done in the case of the chair.

(C) In physics,work is defined as the product of the force applied and the displacement in the direction of the force. The formula is $W = F \cdot s \cdot \cos(\theta)$,where $\theta$ is the angle between the force and the displacement.
If there is no component of force parallel to the direction of motion,then $\theta = 90^\circ$. Since $\cos(90^\circ) = 0$,the work done $W$ becomes $0$.
Therefore,no work is done when there is no component of force parallel to the direction of motion.

(C) 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 vectors.
When a force retards (opposes) the motion of a body,the force acts in the direction opposite to the displacement.
Therefore,the angle $\theta$ between the force and the displacement is $180^{\circ}$.
Since $\cos(180^{\circ}) = -1$,the work done is $W = F \cdot s \cdot (-1) = -F \cdot s$.
Thus,the work done is negative.

(D) The work done $(W)$ by a force $(F)$ on a body is given by the formula $W = F \cdot s \cdot \cos(\theta)$,where $s$ is the displacement and $\theta$ is the angle between the force and the displacement.
For the work done to be positive,$\cos(\theta)$ must be positive,which occurs when $0^\circ \le \theta < 90^\circ$.
This means the body must move in the same direction as the applied force (or at an acute angle to it).
Therefore,if the body moves along the direction of the applied force,the work done is positive.

(A) Work is defined as the product of force and displacement. The $SI$ unit of work is the joule $(J)$,which is equivalent to $1 \text{ N} \cdot \text{m}$.
In the $CGS$ (centimeter-gram-second) system,the unit of force is the dyne and the unit of displacement is the centimeter $(cm)$.
Therefore,the $CGS$ unit of work is the erg,where $1 \text{ erg} = 1 \text{ dyne} \cdot \text{cm}$.
$1 \text{ joule} = 10^7 \text{ ergs}$.

(B) Work done $(W)$ is defined as the product of force $(F)$ and displacement $(s)$ in the direction of the force, given by the formula $W = F \times s$.
One joule $(1 \, J)$ of work is defined as the amount of work done when a force of $1 \, \text{Newton}$ $(1 \, N)$ displaces an object by $1 \, \text{meter}$ $(1 \, m)$ in the direction of the applied force.
Therefore, $1 \, J = 1 \, N \times 1 \, m$.

(B) The statement is False.
In physics, the work done $(W)$ by a force is defined as the product of the force $(F)$ and the displacement $(s)$ in the direction of the force, expressed as $W = F \times s \times \cos(\theta)$.
If a force is applied but produces no motion, the displacement $(s)$ is $0$.
Therefore, the work done is $W = F \times 0 = 0$.
However, the statement says "a force does not work", which implies that work is not being done. Since work is indeed $0$ when there is no motion, the statement is technically true in its physical implication. However, in the context of standard textbook exercises, this is often phrased as "Work is done only when displacement occurs". If the statement implies that the force itself is incapable of doing work, it is False. Given the standard interpretation of this specific question in textbooks, the answer is False because the statement is phrased as a general rule about the force itself rather than the outcome.

(A) The work done $(W)$ is given by the formula $W = Fs \cos \theta$,where $F$ is the force,$s$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
When $\theta = 90^{\circ}$,$\cos 90^{\circ} = 0$.
Therefore,$W = Fs \times 0 = 0$.
Since the work done can be positive,negative,or zero,the minimum value of work done (in terms of magnitude) occurs when the force is perpendicular to the displacement,resulting in zero work.

(FALSE) The statement is False.
Work done by a force is defined as the product of the force applied and the displacement in the direction of the force $(W = F \cdot s \cdot \cos \theta)$.
The rate at which work is done is defined as power $(P = W / t)$.
Therefore,work done depends only on the magnitude of force and displacement,not on the time taken or the speed at which the work is performed.