Question

Write detailed answers?

a. Explain the difference between potential energy and kinetic energy.

b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.

c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.

d. Determine the amount of work done when an object is displaced at an angle of 30^o with respect to the direction of the applied force.

e. If an object has 0 momentum, does it have kinetic energy? Explain your answer.

f. Why is the work done on an object moving with uniform circular motion zero?

Answer 1

a.

Potential Energy	Kinetic Energy
It is possessed by a body by virtue of its configuration and position.	It is possessed by a body by virtue of its motion.
eg: A book kept on a shelf has potential energy.	eg: A moving car posses kinetic energy.

b. Let the initial velocity of the object be u. Let an external force be applied on it so that it gets displaced by distance s and its velocity becomes v. In this scenario, the kinetic energy of the moving body is equal to the work that was required to change its velocity from u to v.

Thus, we have the velocity−position relation as:

v² = u² + 2as

or

Where, a is the acceleration of the body during the change in its velocity

Now, the work done on the body by the external force is given by:

W = F × s

F = ma …(ii)

From equations (i) and (ii), we obtain:

If the body was initially at rest (i.e., u = 0), then:

Since kinetic energy is equal to the work done on the body to change its velocity from 0 to v, we obtain:

c. Let the object be at point A i.e. at height h above the surface of Earth as shown in the figure below.

At point A

The object is stationary i.e. its initial velocity, u = 0.

Kinetic energy, $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{v}^2=0$

Potential energy, U = mgh .....(i)

At point B

Let the velocity of the object be $v_B$ and the object has fallen through distance x.

Using third equation of motion, we have

$v_B^2=0+2gx\Rightarrow v_B^2=2gx$

Kinetic energy, $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_B^2=mgx$

Potential energy, U = mg(h-x) ......(ii)

At point C

Let the velocity of the object be $v_C$ and the object has fallen through a distance h.

Using the third equation of motion, we have

$v_C^2=0+2gh\Rightarrow v_C^2=2gh$

Kinetic energy, $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_C^2=mgh$ ......(iii)

Potential energy, U = 0.

From (i) and (iii), we see that the potential energy of the object at point A has transformed to its kinetic energy at point C. Thus, it can be concluded that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.

d. Work done,$W=F\times s\times cos\left(\theta \right)$

Here, $\theta ={30}^0$

Thus,$W=\frac{\sqrt{3}}{2}F\times s$

e. Momentum of an object, p = mv

If p = 0 ⇒v=0

This is because mass of the object can never be 0.

Now, the kinetic energy of the object, $K=\frac{1}{2}m{v}^2$

Since, v = 0 ⇒K=0

Hence, when the object has zero momentum, its kinetic energy is also zero.

f. Work done on an object is given as $W=F\times s\times cos\left(\theta \right)$

In a circular motion, the direction of force acting on the object is radially inward and the direction of motion of the object is tangential to the circular path at every instant of time. Thus, the angle θ between the force vector and displacement vector is always ${90}^0$ i.e. $\theta ={90}^0$.

Hence, $W=F\times s\times cos\left(90\right)=0$.

Hence, the work done on an object moving in uniform circular motion is zero.