Question

2.2.6 quiz: work and power
question 8 of 10
a sled of mass 3 kg slides along the ice. it has an initial speed of 5 m/s but
stops because of friction. how much work is done by friction?
a. 75 j
b. -75 j
c. 0 j
d. -37.5 j

Explanation:

Step1: Apply work - energy theorem

The work - energy theorem states that \(W_{net}=\Delta K = K_{f}-K_{i}\). Since the sled stops, \(v_{f} = 0\), so \(K_{f}=\frac{1}{2}mv_{f}^{2}=0\). The initial kinetic energy \(K_{i}=\frac{1}{2}mv_{i}^{2}\), where \(m = 3\space kg\) and \(v_{i}=5\space m/s\).

Step2: Calculate the work done by friction

\(W_{friction}=K_{f}-K_{i}=0-\frac{1}{2}mv_{i}^{2}\). Substitute \(m = 3\space kg\) and \(v_{i}=5\space m/s\) into the formula: \(W_{friction}=-\frac{1}{2}\times3\times5^{2}=-\frac{1}{2}\times3\times25=- 37.5\space J\)

Answer:

D. - 37.5 J