Question

an object of mass 1 kg that is moving in a straight line on a level surface slows down from 4 m/s to rest. what is the net work done on the object? hints available hint(s) -8 joules 4 joules 8 joules -4 joules submit return to assignment provide feedback

Explanation:

Step1: Recall Work - Energy Theorem

The work - energy theorem states that the net work done on an object, \(W_{net}\), is equal to the change in the kinetic energy of the object, \(\Delta KE\). The formula for kinetic energy is \(KE=\frac{1}{2}mv^{2}\), where \(m\) is the mass of the object and \(v\) is its velocity. So, \(\Delta KE = KE_{final}-KE_{initial}=\frac{1}{2}mv_{final}^{2}-\frac{1}{2}mv_{initial}^{2}\), and \(W_{net}=\Delta KE\).

Step2: Identify Given Values

We are given that the mass of the object, \(m = 1\space kg\), the initial velocity, \(v_{initial}=4\space m/s\), and the final velocity, \(v_{final} = 0\space m/s\) (since the object comes to rest).

Step3: Calculate Initial and Final Kinetic Energy

• Calculate the initial kinetic energy: \(KE_{initial}=\frac{1}{2}mv_{initial}^{2}\). Substituting \(m = 1\space kg\) and \(v_{initial}=4\space m/s\) into the formula, we get \(KE_{initial}=\frac{1}{2}\times1\times(4)^{2}=\frac{1}{2}\times1\times16 = 8\space J\).
• Calculate the final kinetic energy: \(KE_{final}=\frac{1}{2}mv_{final}^{2}\). Substituting \(m = 1\space kg\) and \(v_{final}=0\space m/s\) into the formula, we get \(KE_{final}=\frac{1}{2}\times1\times(0)^{2}=0\space J\).

Step4: Calculate Net Work Done

Using the work - energy theorem \(W_{net}=\Delta KE=KE_{final}-KE_{initial}\). Substituting the values of \(KE_{final}\) and \(KE_{initial}\), we have \(W_{net}=0 - 8=- 8\space J\)? Wait, no, wait. Wait, let's recalculate. Wait, \(v_{initial} = 4\), \(m = 1\). Wait, \(\frac{1}{2}mv^{2}\) when \(v = 4\) is \(\frac{1}{2}\times1\times16 = 8\)? Wait, no, the final velocity is 0, so \(\Delta KE=0-\frac{1}{2}\times1\times4^{2}=0 - 8=- 8\)? But wait, the options have - 8? Wait, no, wait the options: - 8, 4, 8, - 4. Wait, maybe I made a mistake. Wait, no, the work - energy theorem: net work done is change in kinetic energy. Initial KE: \(\frac{1}{2}\times1\times4^{2}=8\space J\), final KE: 0. So change in KE is \(0 - 8=- 8\space J\). But wait, the options have - 8 as an option. Wait, but let me check again. Wait, maybe the question is about the magnitude? No, the work done should be negative because the object is slowing down, so the net force is opposite to the displacement. So the net work done is \(- 8\space J\)? Wait, but let me check the calculation again.

Wait, \(KE=\frac{1}{2}mv^{2}\), so initial \(KE=\frac{1}{2}\times1\times4^{2}=8\space J\), final \(KE = 0\). So \(\Delta KE=0 - 8=- 8\space J\). So the net work done is \(- 8\space J\). But wait, the options have - 8 as the first option. Wait, but maybe I messed up the sign. Wait, the work - energy theorem says \(W_{net}=\Delta KE\). If the object is slowing down, its kinetic energy decreases, so \(\Delta KE\) is negative, so \(W_{net}\) is negative. So the answer should be - 8 Joules. Wait, but let me check the options again. The first option is - 8 Joules.

Wait, no, wait, maybe I made a mistake in the velocity. Wait, the initial velocity is 4 m/s, mass 1 kg. So initial KE is 8 J, final KE is 0. So change in KE is - 8 J, so net work done is - 8 J. So the correct option is - 8 Joules.

Answer:

-8 Joules (the option corresponding to - 8 Joules)