Question

A 25 kg cart is moving at 2.0 m/s when a 5.0 N force is applied in the direction it is moving for a distance of 6.0 m. What is the kinetic energy of the cart at the end of the 5.0 m?

Answer 1

The initial kinetic energy of the cart is:

`K_i = (1)/(2)mv^2 = (1)/(2)(25 kg)(2m/s)^2=50 J`
where m is the mass of the cart and v its initial speed.

Then, the force F is applied, in the direction of motion of the cart.The work done by this force is:

`W=Fd=(5 N)(6.0 m)=30 J`

For the work-energy theorem, the work done by the force is equal to the variation of kinetic energy of the cart:

`K_f - K_i = W`
And so, the final kinetic energy of the cart is

`K_f = K_i + W=50 J+30 J=80 J`