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 change in kinetic energy. The initial kinetic energy is $\frac{1}{2} \times 1000 \, \text{kg} \times (26.1 \, \text{m/s})^2 = 3.41 \times 10^5 \, \text{J}$. Since the car comes to rest, the final kinetic energy is 0, so the work done is $3.41 \times 10^5 \, \text{J}$.

How Much Work Must Be Done by Frictional Forces in Slowing