Question

A 1000 kg automobile is traveling at an initial speed of 20 m/s. It is brought to a complete stop in 5 s over a distance of 50 m. What is the work done in stopping the automobile

Answer 1

Answer:

W = -2*10⁵ J

Explanation:

• Assuming no friction present, we can find the work done by an external force stopping the car, applying the work-energy theorem.
• This theorem says that the total work done on one object by an external net force, is equal to the change in the kinetic energy of the object.
• If the automobile is brought to a complete stop, we can find the change of the kinetic energy as follows:

        $\Delta K = K_{f} - K_{0} = 0 - \frac{1}{2} * m* v_{0} ^{2} \\\\ \Delta K = -\frac{1}{2}*1000kg*(20 m/s)^{2} = -200000 J= -2e5 J$

• So, the total work done in stopping the automobile, is -2*10⁵ J. The minus sign stems from the fact that the force and the displacement have opposite directions.