Question

question 18 of 27
a car of mass 500 kg is moving at a speed of 1.2 m/s. a man pushes the car, increasing the speed to 2 m/s. how much work did the man do?
a. 360 j
b. 1360 j
c. 640 j
d. 1000 j

Explanation:

Step1: Recall work - energy theorem

The work done on an object is equal to the change in its kinetic energy, $W=\Delta K = K_f - K_i$. The formula for kinetic energy is $K=\frac{1}{2}mv^{2}$, where $m$ is the mass of the object and $v$ is its speed.

Step2: Calculate initial kinetic energy

Given $m = 500$ kg and $v_i=1.2$ m/s. Using the kinetic - energy formula $K_i=\frac{1}{2}mv_i^{2}=\frac{1}{2}\times500\times(1.2)^{2}= \frac{1}{2}\times500\times1.44 = 360$ J.

Step3: Calculate final kinetic energy

Given $v_f = 2$ m/s and $m = 500$ kg. Using the kinetic - energy formula $K_f=\frac{1}{2}mv_f^{2}=\frac{1}{2}\times500\times(2)^{2}=\frac{1}{2}\times500\times4 = 1000$ J.

Step4: Calculate the work done

$W=K_f - K_i=1000 - 360=640$ J.

Answer:

C. 640 J