Question

was moving at a rate of 8 m/s, and then sped up to 10 m/s. if the has a mass of 120 kg, how much work was needed to increase his (hint: use the work - kinetic energy theorem.)
a 3240 j
b 1220 j
c 2160 j
d 6220 j

Explanation:

Step1: Recall the work - kinetic energy theorem

The work - kinetic energy theorem states that \(W=\Delta K = K_{f}-K_{i}\), where the kinetic energy formula is \(K=\frac{1}{2}mv^{2}\).

Step2: Calculate the initial kinetic energy \(K_{i}\)

Given \(m = 120\space kg\) and \(v_{i}=8\space m/s\), then \(K_{i}=\frac{1}{2}\times120\times8^{2}=3840\space J\).

Step3: Calculate the final kinetic energy \(K_{f}\)

Given \(m = 120\space kg\) and \(v_{f}=10\space m/s\), then \(K_{f}=\frac{1}{2}\times120\times10^{2}=6000\space J\).

Step4: Calculate the work done \(W\)

Using \(W = K_{f}-K_{i}\), substitute the values: \(W=6000 - 3840=2160\space J\).

Answer:

C. \(2160\space J\)