Question

A 15.0 kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of 10.0 N acts on the cart, and its velocity become 3.20 m/s.
a) What is the change in kinetic energy?
b) How much work was done on the cart?
c) How far did the cart move while the force acted?

Answer 1

Part a). We know that kinetic energy is defined as: \(1/2 mv^2\). Therefore, the change in kinetic energy can be expressed as: \[\Delta KE = KE_f - KE_i = {1 \over 2} m v_f^2 - {1 \over 2} m v_i^2\] (The change will be negative in this case.)

Part b). The work done can be expressed in terms of the change in energy of the cart. \[W = \Delta KE + \Delta PE\]Since the hallway is level, \(\Delta PE = 0\). The work is simply the change in kinetic energy.

Part c). Since the force is constant, the acceleration it causes will also be constant. Therefore, \[\Delta x = \left ( v_f + v_i \over 2 \right) t\]To find \(t\), we can use the following equation: \[v_f = at + v_i\]