Chapter 6, Problem 18P

How much work must be done to stop a 925-kg car travelling at 95 km/h?

To determine

Work required to stop a car.

Answer to Problem 18P

Solution: $3.22\times {10}^5\text{J}$

Explanation of Solution

Given:Mass of the car, $m=925\text{ kg}$

Velocity of electron, $v=95\text{ km/h}$

Formula Used:

According to work energy theorem, work done by the force equal to change in kinetic energy of particle:

  $\begin{array}{l}W=\Delta K.E\\ W=K.E_{final}-K.E_{initial}\end{array}$

Here final kinetic energy is zero

  $W=-K.E_{initial}$

  $W=-\frac{1}{2}m{v}^2\mathrm{....}(1)$

Calculations:

  $\begin{array}{c}v=95\times \frac{1000}{3600}\\ =26.388\text{ m/s}\end{array}$

Substitute this value in equation(1)

  $\begin{array}{l}$ W=-\frac{1}{2}925\times {(26.388)}^2$W=-3.22\times {10}^5\text{ J}\end{array}$

Work required to stop the car is $3.22\times {10}^{\text{5}}\text{J}$ .

Conclusion: $3.22\times {10}^{\text{5}}\text{J}$ work required to stop a car mass $925\text{ kg}$ which is travelling at $95\text{ km/h}.$