How much work must be done by frictional forces in slowing a 1000-kg car from $26.1 \mathrm {~m} / \mathrm { s }$ to rest? A)3.41 × 10⁵ J B)2.73 × 10⁵ J C)4.09 × 10⁵ J D)4.77 × 10⁵ J

Question

Quizplus · Accepted Answer

The work done by frictional forces is equal to the initial kinetic energy of the car, which is $\frac{1}{2}mv^2 = \frac{1}{2}(1000 \, \text{kg})(26.1 \, \text{m/s})^2 = 3.41 \times 10^5 \, \text{J}$.

How Much Work Must Be Done by Frictional Forces in Slowing