Question

A man is standing on a cart of mass double the mass of the man. Initially the cart is at rest. Now the man jumps horizontally with a velocity u relative to the cart. Then calculate the work done by the man during the process of jumping.

Answer 1

Answer:

The overall momentum of a system of separate objects (such as this man and his cart) is always constant. And, because the individual is moving, the work done will be equal to the system's kinetic energy.

KE=mu22, where m is the object's mass and u is its velocity. The kinetic energy owing to the object's motion is denoted by KE.

Explanation:

• A man and a cart form the basis of our two-body system. In this situation, we are asked to calculate the total work done or the kinetic energy created by the guy moving at the same speed as the cart. The following is a list of the data that has been submitted to us:
• The man's weight is m.
• The cart's mass is 2m.
• Let's pretend we have the following variables:
• The man's speed is vm.
• The cart's speed is v.
• When in motion, vm+v=u, according to the query.
• This is the same as: vm=uv
• We already know that the system's momentum will remain constant:
• ⇒mvm=2mv
• ⇒m(u−v)=2mv
• When we solve for the cart's velocity, we get: uv=2v
• ⇒v=u3
• The velocity of the guy will be vm=uv=uu3=2u3 as a result of this.
• KE=mvm22+2mv22 KE=m(2u3)22+m(u3)2 KE=m(2u3)22+m(u3)2 KE=m(2u3)22+m(u3)2 KE=m(2u3)22+m(u3)2 KE=m(2u3)22+m
• When we open the brackets and solve the problem further, we get:
• ⇒KE=m×4u22×9+mu29=mu2(29+19)
• ⇒KE=mu2×39=mu23