How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest?
A) $3.41 \times 10 ^ { 5 } \mathrm {~J}$
B) $2.73 \times 10 ^ { 5 } \mathrm {~J}$
C) $4.77 \times 10 ^ { 5 } \mathrm {~J}$
D) $4.09 \times 10 ^ { 5 } \mathrm {~J}$

Question

How much work must be done by frictional forces in slowing a  1000-kg car from  26.1 m/s to rest?
A) $3.41 \times 10 ^ { 5 } \mathrm {~J}$
B) $2.73 \times 10 ^ { 5 } \mathrm {~J}$
C) $4.77 \times 10 ^ { 5 } \mathrm {~J}$
D) $4.09 \times 10 ^ { 5 } \mathrm {~J}$

Quizplus · Accepted Answer

The work done by frictional forces in slowing the car to rest can be calculated using the work-energy principle, which states that the work done is equal to the change in kinetic energy. The initial kinetic energy ($KE_i$) of the car is given by $\frac{1}{2}mv^2$, where $m = 1000 \, \text{kg}$ is the mass of the car and $v = 26.1 \, \text{m/s}$ is its initial velocity. The final kinetic energy ($KE_f$) is 0 because the car comes to rest. Thus, the work done by friction ($W$) is equal to the change in kinetic energy ($KE_f - KE_i$):$$W = KE_f - KE_i = 0 - \frac{1}{2} \times 1000 \, \text{kg} \times (26.1 \, \text{m/s})^2$$$$W = - \frac{1}{2} \times 1000 \times 680.41$$$$W = - 340205 \, \text{J}$$The negative sign indicates that the work is done against the motion of the car. However, when considering the magnitude of work done (which is what the question effectively asks for), we get:$$|W| = 3.41 \times 10^5 \, \text{J}$$Therefore, the correct answer is $3.41 \times 10^5 \, \text{J}$.

How Much Work Must Be Done by Frictional Forces in Slowing