Question

The masses m1 and m2 are released from rest using work energy theorem find out velocity of the blocks when they move a distance 'x'

Answer 1

ANSWER:

The velocities exhibited by the blocks are 2x/t.

EXPLANATION:

According to "work energy theorem", the change in the kinetic energy of a body will be equal to the work done on the body or by the body.

So

$k_{f}-k_{i}=W$

Here ki and kf are the initial and final kinetic energy exerted by the body and W is the work done by the body. We know that work done can be calculated by dot product of force acting on the body and the distance travelled by the body after exertion of force. Thus

$k_{f}-k_{i}=F, d$

As the blocks are released that means the initial velocity of the blocks will be zero, so the initial kinetic energy of the blocks will also be zero.

So the final velocity of block with mass m1 when it moved a distance of x will be as follows:

$k_{f}=F \cdot x$

As according to Newton's second law, force is directly proportional to the "mass of the object" and the acceleration exerted by the object. So in that situation:

$k_{f}=m_{1} a \cdot x$

$\frac{1}{2} m_{1} v_{1}^{2}=m_{1} \cdot \frac{v_{1}}{t} \cdot x$

$v_{1}=\frac{2 x}{t}$

Similarly, the block with mass as m2 will also exhibit similar velocity as 2x/t.