Question

a 60.0 kg object is moving east at 8.00 m/s, and then slows down to 4.00 m/s. how much work was done?
-1,440 j
-480 j
1,440 j
2,880 j

Explanation:

Step1: Recall work - energy theorem

The work - energy theorem states that $W=\Delta K = K_f - K_i$, where $W$ is the work done, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy. The formula for kinetic energy is $K=\frac{1}{2}mv^{2}$, with $m$ being the mass and $v$ the velocity.

Step2: Calculate the initial kinetic energy

Given $m = 60.0\ kg$ and $v_i=8.00\ m/s$, then $K_i=\frac{1}{2}mv_i^{2}=\frac{1}{2}\times60.0\times(8.00)^{2}= \frac{1}{2}\times60.0\times64.0 = 1920\ J$.

Step3: Calculate the final kinetic energy

Given $m = 60.0\ kg$ and $v_f = 4.00\ m/s$, then $K_f=\frac{1}{2}mv_f^{2}=\frac{1}{2}\times60.0\times(4.00)^{2}=\frac{1}{2}\times60.0\times16.0 = 480\ J$.

Step4: Calculate the work done

$W=\Delta K=K_f - K_i=480 - 1920=- 1440\ J$.

Answer:

A. -1,440 J