Kinetic and Potential Energy Questions in English

(N/A) The kinetic energy $(K)$ of a body of mass $m$ moving with velocity $v$ is given by the formula: $K = \frac{1}{2}mv^2$.
$(i)$ If the mass is doubled $(m' = 2m)$ and velocity remains constant $(v' = v)$:
$K' = \frac{1}{2}(2m)v^2 = 2 \times (\frac{1}{2}mv^2) = 2K$.
The kinetic energy becomes double the original value.
$(ii)$ If the velocity is doubled $(v' = 2v)$ and mass remains constant $(m' = m)$:
$K' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times (\frac{1}{2}mv^2) = 4K$.
The kinetic energy becomes four times the original value.
$(iii)$ If the mass is doubled $(m' = 2m)$ and velocity is reduced to half $(v' = v/2)$:
$K' = \frac{1}{2}(2m)(v/2)^2 = \frac{1}{2}(2m)(v^2/4) = \frac{1}{2} \times (\frac{1}{2}mv^2) = K/2$.
The kinetic energy becomes half of the original value.

(B) The relationship between momentum $(p)$,mass $(m)$,and kinetic energy $(K)$ is given by the formula $p = \sqrt{2mK}$.
Since the kinetic energy $(K)$ is the same for both bodies,the momentum is directly proportional to the square root of the mass $(p \propto \sqrt{m})$.
Therefore,the body with the greater mass (the heavy body) will have a greater momentum compared to the lighter body.

(A) The relationship between kinetic energy $(K)$,momentum $(p)$,and mass $(m)$ is given by the formula $K = \frac{p^2}{2m}$.
Since the momentum $(p)$ is the same for both bodies,the kinetic energy is inversely proportional to the mass $(K \propto \frac{1}{m})$.
Therefore,the body with the smaller mass (the lighter body) will have a larger denominator,resulting in a greater value for kinetic energy.
Thus,the lighter body has more kinetic energy than the heavier body.

(B) The kinetic energy $(KE)$ of an object is given by the formula: $KE = \frac{1}{2}mv^2$.
$1$. If we double the mass $(m' = 2m)$,the new kinetic energy becomes: $KE' = \frac{1}{2}(2m)v^2 = 2 \times KE$. Thus,doubling the mass doubles the kinetic energy.
$2$. If we double the velocity $(v' = 2v)$,the new kinetic energy becomes: $KE'' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times KE$. Thus,doubling the velocity increases the kinetic energy by four times.
Since $4 \times KE > 2 \times KE$,doubling the velocity has a greater effect on the kinetic energy of an object.

(B) When water is at a certain height,it possesses gravitational potential energy.
As the water falls,this gravitational potential energy is converted into kinetic energy.
Upon reaching the bottom,the water molecules collide with each other and with the rocks,causing this kinetic energy to dissipate as heat energy.
Consequently,the water at the bottom of the waterfall is warmer than the water at the top.

(A) When a coil spring is compressed,an external force is applied to displace the spring from its equilibrium position.
According to the work-energy theorem,the work done on the spring by the external force is stored within the spring in the form of elastic potential energy.
The formula for elastic potential energy is $U = \frac{1}{2}kx^2$,where $k$ is the spring constant and $x$ is the displacement.
As the spring is compressed,the displacement $x$ increases,which leads to an increase in the elastic potential energy stored in the spring.

(B) The gravitational potential energy $(PE)$ of an object is given by the formula $PE = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the height of the center of mass of the object from the reference level (ground).
Since $m$ and $g$ are constant for a person,the potential energy depends directly on the height $(h)$ of the center of mass.
When a person is lying on the ground,the height of their center of mass from the ground is the minimum compared to standing,sitting on the ground,or sitting on a chair.
Therefore,the potential energy is minimum when the person is lying on the ground.

(D) The kinetic energy $(K.E.)$ of an object of mass $m$ moving with velocity $v$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.
If the speed is doubled,the new velocity becomes $v' = 2v$.
The new kinetic energy $(K.E.')$ will be: $K.E.' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times (\frac{1}{2}mv^2)$.
Therefore,$K.E.' = 4 \times K.E.$
This shows that the kinetic energy increases $4$ times when the speed is doubled.

(A) The gravitational potential energy $(PE)$ of an object of mass $m$ at a height $h$ above the ground is given by the formula $PE = mgh$.
When the stone falls through a vertical height $d$,its initial height is $h_i$ and its final height is $h_f = h_i - d$.
The change in gravitational potential energy is $\Delta PE = PE_i - PE_f$.
$\Delta PE = mgh_i - mg(h_i - d)$.
$\Delta PE = mgh_i - mgh_i + mgd$.
$\Delta PE = mgd$.
Therefore,the decrease in gravitational potential energy is $mgd$.

(C) The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2} mv^2$,where $m$ is the mass and $v$ is the velocity.
Given: $KE = 1 \, J$,$m = 1 \, kg$.
Substituting the values into the formula: $1 = \frac{1}{2} \times 1 \times v^2$.
$v^2 = 2$.
$v = \sqrt{2} \approx 1.414 \, m/s$.
Therefore,the speed is approximately $1.4 \, m/s$.

(D) The formula for gravitational potential energy $(PE)$ is given by $PE = mgh$,where:
$m$ is the mass of the object $(1 \, kg)$,
$g$ is the acceleration due to gravity (approximately $9.8 \, m/s^2$),
$h$ is the height above the ground.
Given $PE = 1 \, J$,we have:
$1 = 1 \times 9.8 \times h$
$h = 1 / 9.8 \, m$
$h \approx 0.102 \, m$.
Therefore,the correct height is $0.102 \, m$.

(A) The potential energy $(PE)$ of an object at a height $(h)$ is given by the formula: $PE = m \cdot g \cdot h$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the height.
Given: Mass $(m)$ = $5\,kg$,Height $(h)$ = $5\,m$,Acceleration due to gravity $(g)$ $\approx 10\,m/s^2$.
The loss of potential energy as the object falls to the ground is equal to the initial potential energy at height $h$.
$PE = 5\,kg \times 10\,m/s^2 \times 5\,m = 250\,J$.
Therefore,the loss of potential energy is $250\,J$.

(C) The kinetic energy $(K)$ of a particle of mass $(m)$ moving with speed $(v)$ is given by the formula: $K = \frac{1}{2}mv^2$.
If the speed is doubled,the new speed becomes $v' = 2v$.
The new kinetic energy $(K')$ will be: $K' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(2v)^2$.
$K' = \frac{1}{2}m(4v^2) = 4 \times (\frac{1}{2}mv^2) = 4K$.
Therefore,the kinetic energy becomes four times the original kinetic energy.

(TRUE) The statement is True.
Kinetic energy $(KE)$ is given by the formula: $KE = \frac{1}{2}mv^2$,where $m$ is mass and $v$ is velocity.
If the velocity is halved,the new velocity $v' = \frac{v}{2}$.
The new kinetic energy $KE'$ will be: $KE' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{2}m(\frac{v^2}{4}) = \frac{1}{4} (\frac{1}{2}mv^2) = \frac{1}{4} KE$.
Therefore,the kinetic energy becomes $\frac{1}{4}^{th}$ of the original kinetic energy.

(TRUE) The statement is True.
When the string of a bow is pulled,work is done on the bow,which is stored as elastic potential energy in the deformed bow. When the arrow is released,this stored elastic potential energy is converted into the kinetic energy of the arrow,causing it to move forward.

(TRUE) The statement is True.
The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2}mv^2$,where $m$ is the mass of the object and $v$ is its speed (or velocity).
Since the formula requires only the mass $(m)$ and the speed $(v)$ of the object to calculate the kinetic energy,knowing these two values allows us to determine the kinetic energy of the object.