Mix Example - WORK AND ENERGY Questions in English

(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 boy has to overcome the force of gravity to climb the stairs.
Force of gravity on the boy,$F = m \times g = 40 \, kg \times 9.8 \, m s^{-2} = 392 \, N$.
Total height (distance) covered,$S = 50 \times 10 \, cm = 500 \, cm = 5 \, m$.
$(i)$ Work done by the boy in climbing,$W = F \times S = 392 \, N \times 5 \, m = 1960 \, J$.
$(ii)$ Power developed,$P = \frac{W}{t} = \frac{1960 \, J}{5 \, s} = 392 \, W$.

(B) The work done $W$ in lifting a mass $m$ to a height $h$ against the gravitational force is given by $W = mgh$.
Power $P$ is defined as the rate of doing work,$P = \frac{W}{t}$.
Given values are: $m = 10^{5} \text{ kg}$,$g = 10 \text{ m s}^{-2}$,$h = 360 \text{ m}$,and $t = 1 \text{ hour} = 3600 \text{ s}$.
Substituting these values into the formula:
$P = \frac{10^{5} \times 10 \times 360}{3600}$
$P = \frac{10^{6} \times 360}{3600}$
$P = 10^{5} \text{ W}$.
Thus,the power required is $10^{5} \text{ W}$.

(C) Given: Power of the motor $P_{total} = 100\, W$, Time $t = 1\, \text{minute} = 60\, \text{seconds}$.
The power effectively used for stirring the water is $50\%$ of the total power supplied.
Power used for stirring $P_{stir} = 50\% \text{ of } 100\, W = \frac{50}{100} \times 100 = 50\, W$.
Work done on the water is calculated using the formula: $\text{Work} = \text{Power} \times \text{Time}$.
$\text{Work} = 50\, W \times 60\, \text{s} = 3000\, J$.
Therefore, the work done on the water in one minute is $3000\, J$.

(C) Given: Mass $m = 10 \, kg$,Height $h = 50 \, cm = 0.5 \, m$,Acceleration due to gravity $g = 10 \, m s^{-2}$.
$(a)$ Potential energy $(PE)$ just before dropping is calculated as $PE = mgh = 10 \times 10 \times 0.5 = 50 \, J$.
$(b)$ According to the law of conservation of energy,the total mechanical energy remains constant. Just before touching the ground,the potential energy is converted into kinetic energy $(KE)$. Therefore,$KE = PE = 50 \, J$.
$(c)$ Using the formula for kinetic energy,$KE = \frac{1}{2}mv^2$,we can find the velocity $v$ as $v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 50}{10}} = \sqrt{10} \approx 3.16 \, m s^{-1}$.

(D) Energy consumed is given by $E = P \times t$.
Energy consumed by $4$ bulbs per day $= 4 \times 40 \, W \times 5 \, hrs = 800 \, Wh$.
Energy consumed by $4$ tube lights per day $= 4 \times 60 \, W \times 5 \, hrs = 1200 \, Wh$.
Energy consumed by $TV$ per day $= 100 \, W \times 6 \, hrs = 600 \, Wh$.
Energy consumed by washing machine per day $= 400 \, W \times 3 \, hrs = 1200 \, Wh$.
Total energy consumed per day $= 800 + 1200 + 600 + 1200 = 3800 \, Wh$.
Since April has $30$ days, total energy consumed in April $= 3800 \, Wh \times 30 = 114000 \, Wh$.
Converting to units $(kWh)$: $114000 \, Wh / 1000 = 114 \, kWh$ (or $114$ units).
Total electricity bill $= 114 \, \text{units} \times ₹ 1.80 / \text{unit} = ₹ 205.20$.

(A) Given: Mass $m = 120 \, kg$,initial velocity $u = 25 \, m s^{-1}$,final velocity $v = 40 \, m s^{-1}$.
According to the work-energy theorem,the work done $W$ on an object is equal to the change in its kinetic energy.
$W = \Delta KE = \frac{1}{2} m (v^2 - u^2)$
Substituting the values:
$W = \frac{1}{2} \times 120 \times (40^2 - 25^2)$
$W = 60 \times (1600 - 625)$
$W = 60 \times 975$
$W = 58500 \, J$
Thus,the work done is $58500 \, J$.

(B) Initial potential energy of the ball at height $h = 10 \, m$ is $E = mgh$.
After striking the ground,the energy reduces by $40 \%$,which means the ball retains $60 \%$ of its initial energy.
Energy retained $= 0.60 \times mgh$.
When the ball bounces back to a new height $h'$,its potential energy at that height will be $mgh'$.
Equating the retained energy to the new potential energy: $mgh' = 0.60 \times mgh$.
Canceling $mg$ from both sides,we get $h' = 0.60 \times h$.
Substituting $h = 10 \, m$,we get $h' = 0.60 \times 10 \, m = 6 \, m$.
Therefore,the ball bounces back to a height of $6 \, m$.

(B) Given: Mass $m = 5\, kg$,Initial velocity $u = 0\, m/s$,Force $F = 20\, N$,Time $t = 10\, s$.
Step $1$: Calculate acceleration using Newton's second law of motion,$F = ma$.
$a = F / m = 20 / 5 = 4\, m/s^2$.
Step $2$: Calculate the final velocity $v$ using the first equation of motion,$v = u + at$.
$v = 0 + (4 \times 10) = 40\, m/s$.
Step $3$: Calculate the kinetic energy $KE$ using the formula $KE = 1/2 mv^2$.
$KE = 1/2 \times 5 \times (40)^2 = 1/2 \times 5 \times 1600 = 5 \times 800 = 4000\, J$.

(N/A) Given: Mass $m = 10 \, kg$,Height $h = 20 \, m$,Acceleration due to gravity $g = 10 \, m s^{-2}$.
$1$. Kinetic Energy $(KE)$ at the ground:
According to the law of conservation of energy,the potential energy $(PE)$ at the top is converted into kinetic energy $(KE)$ at the ground.
$PE = mgh = 10 \times 10 \times 20 = 2000 \, J$.
Therefore,$KE = 2000 \, J$.
$2$. Speed $(v)$ just before hitting the ground:
Using the formula $KE = \frac{1}{2} mv^2$:
$2000 = \frac{1}{2} \times 10 \times v^2$
$2000 = 5 \times v^2$
$v^2 = 400$
$v = \sqrt{400} = 20 \, m s^{-1}$.

(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.

(N/A) Given: Mass of car $m_{c} = 750\, kg$,Velocity $v = 54\, km h^{-1} = 54 \times \frac{5}{18} = 15\, m s^{-1}$.
Initial Kinetic Energy $(KE_{i})$ = $\frac{1}{2} m_{c} v^{2} = \frac{1}{2} \times 750 \times (15)^{2} = 375 \times 225 = 84375\, J$.
When a passenger of mass $50\, kg$ sits in the car,the new total mass $M = 750 + 50 = 800\, kg$.
New Kinetic Energy $(KE_{f})$ = $\frac{1}{2} M v^{2} = \frac{1}{2} \times 800 \times (15)^{2} = 400 \times 225 = 90000\, J$.

(A) According to the Law of Conservation of Energy,the total mechanical energy of an object remains constant in the absence of air resistance.
At the maximum height $h$,the ball possesses potential energy equal to $mgh$ and its kinetic energy is $0$.
As the ball falls,its potential energy is converted into kinetic energy.
Just before touching the ground,the entire potential energy is converted into kinetic energy.
Therefore,the kinetic energy of the ball just before touching the ground is equal to $mgh$.

(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 $(W) = m \times g \times h$
$W = 45 \, kg \times 10 \, m/s^2 \times 10 \, m$
$W = 4500 \, J$.

(A) Work done by each child is calculated using the formula $W = mgh$.
Given $m = 20 \, kg$,$g = 9.8 \, m/s^2$,and $h = 10 \, m$.
Work done $= 20 \times 9.8 \times 10 = 1960 \, J$.
Since $m$,$g$,and $h$ are the same for both children,the work done is equal.
Power is defined as the rate of doing work,$P = W / t$.
For child $A$: $P_A = 1960 / 10 = 196 \, W$.
For child $B$: $P_B = 1960 / 20 = 98 \, W$.
Since $196 \, W > 98 \, W$,child $A$ has more power.

(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.

(D) compressed spring stores elastic potential energy due to its deformed state.
When the spring is dissolved in an acid,the chemical bonds holding the spring in its compressed state are broken.
The stored elastic potential energy is released and converted into the kinetic energy of the acid molecules.
This increase in the kinetic energy of the molecules results in a rise in the temperature of the acid solution.
Therefore,the potential energy is ultimately converted into thermal energy.

(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.

(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.

(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.

(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.

(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 $1.4 \, m/s$.

(C) When objects are dropped from a height,they undergo free fall under the influence of gravity.
In the absence of air resistance,the acceleration of an object in free fall is equal to the acceleration due to gravity $(g)$,which is approximately $9.8 \, m/s^2$.
This acceleration is independent of the mass,size,or material of the object.
Since both spheres are dropped simultaneously from the same height,at any given point (including $10 \, m$ above the ground),they will have the same velocity and the same acceleration $(g)$.
Potential energy $(PE = mgh)$ depends on mass,so it will be different for the two spheres.
Momentum $(p = mv)$ depends on mass,so it will be different.
Kinetic energy $(KE = 1/2 mv^2)$ depends on mass,so it will be different.
Therefore,the only quantity that remains the same for both spheres is their acceleration.

(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 $250\,J$.

(B) The work done $(W)$ by the weightlifter is equal to the change in potential energy,given by $W = mgh$.
Here,mass $(m)$ = $240 \,kg$,height $(h)$ = $2.5 \,m$,and acceleration due to gravity $(g)$ = $9.8 \,m/s^2$.
$W = 240 \times 9.8 \times 2.5 = 5880 \,J$.
The time taken $(t)$ = $3 \,s$.
Average power $(P)$ is defined as the rate of doing work,$P = W / t$.
$P = 5880 / 3 = 1960 \,W$.
Therefore,the average power is $1960 \,W$.

(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, $1 \, J = 1 \, N \times 1 \, m$.

(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.
This is a fundamental principle known as the Law of Conservation of Energy.
According to this law,energy can neither be created nor destroyed,only transformed from one form to another.
Therefore,the total energy of an isolated system remains constant,meaning the amount of energy lost by one form is exactly gained by the other form.

(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.

(B) The statement is False.
Kilowatt hour $(kWh)$ is a unit of energy,not power.
Power is measured in Watts $(W)$ or Kilowatts $(kW)$,while energy is measured in Kilowatt hours $(kWh)$.

(FALSE) The statement is False.
In physics,work is defined as the product of force and displacement,and energy is defined as the capacity to do work.
Both work and energy are measured in the same $SI$ unit,which is the Joule $(J)$.

(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.
According to the law of conservation of energy,energy can neither be created nor destroyed,only transformed from one form to another.
When a body falls and hits the ground,its kinetic energy is converted into other forms of energy such as sound energy,heat energy,and potential energy of deformation (if the body or ground is deformed).
Therefore,the total energy of the system remains conserved,and the principle is not violated.

(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.

(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.

(A) The statement is True.
Power is defined as the rate at which work is done or the rate at which energy is transferred.
Mathematically, $Power (P) = \frac{Work (W)}{Time (t)}$.
Its $SI$ unit is the Watt $(W)$.

(A) The statement is $True$.
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 circular motion,the centripetal force acts towards the center of the circle,while the displacement (tangential velocity) is always perpendicular to the radius.
Since the angle $\theta$ between the centripetal force and the displacement is $90^{\circ}$,the work done is $W = F \cdot s \cdot \cos(90^{\circ})$.
Because $\cos(90^{\circ}) = 0$,the work done by the centripetal force is $0$.

(FALSE) The statement is False.
The $SI$ unit of work is the joule $(J)$,which is defined as the work done when a force of $1$ newton displaces an object by $1$ meter.
Watt $(W)$ is the $SI$ unit of power,which is defined as the rate of doing work or the rate of energy transfer ($1$ watt = $1$ joule per second).

(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.