Question

question 4 of 10
an engine performs 2700 j of work on a scooter. the scooter and rider have
a combined mass of 150 kg. if the scooter started at rest, what is the speed
of the bike after the work is performed?
o a. 6 m/s
o b. 8 m/s
o c. 7 m/s
o d. 5 m/s

Explanation:

Step1: Use work - energy theorem

The work - energy theorem states that \(W=\Delta K\). Since the scooter starts from rest (\(v_{0} = 0\)), the initial kinetic energy \(K_{0}=\frac{1}{2}mv_{0}^{2}=0\). The work done \(W\) is equal to the final kinetic energy \(K_{f}\). The formula for kinetic energy is \(K=\frac{1}{2}mv^{2}\), where \(W = K_{f}\), \(m = 150\space kg\) and \(W=2700\space J\).

Step2: Solve for velocity \(v\)

From \(W=\frac{1}{2}mv^{2}\), we can solve for \(v\). Rearranging the formula gives \(v^{2}=\frac{2W}{m}\). Substitute \(W = 2700\space J\) and \(m = 150\space kg\) into the formula: \(v^{2}=\frac{2\times2700}{150}\).

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Take the square root of both sides: \(v=\sqrt{36}=6\space m/s\)

Answer:

A. \(6\space m/s\)