Question

a 1,500 kg cars speed changes from 30 m/s to 15 m/s after the brakes are applied. calculate the work done onto the car from the brakes. (1 point) -506,250 j 506,250 j -168,750 j 168,750 j

Explanation:

Step1: Recall the work - energy theorem

The work - energy theorem states that \(W=\Delta K = K_{f}-K_{i}\), where \(K=\frac{1}{2}mv^{2}\) is the kinetic energy, \(m\) is the mass of the object, and \(v\) is the speed of the object.

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

Given \(m = 1500\space kg\) and \(v_{i}=30\space m/s\), then \(K_{i}=\frac{1}{2}mv_{i}^{2}=\frac{1}{2}\times1500\times30^{2}=675000\space J\)

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

Given \(v_{f} = 15\space m/s\), then \(K_{f}=\frac{1}{2}mv_{f}^{2}=\frac{1}{2}\times1500\times15^{2}=168750\space J\)

Step4: Calculate the work done \(W\)

\(W=K_{f}-K_{i}=168750 - 675000=- 506250\space J\)

Answer:

\(-506,250\space J\)