a car of mass 550 kg is moving at a speed of 1.5 m/s. a man pushes the car, increasing the speed to 3.2 m/s…

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 the initial kinetic energy

Given $m = 550$ kg and $v_i=1.5$ m/s. Using the kinetic - energy formula $K_i=\frac{1}{2}mv_i^{2}=\frac{1}{2}\times550\times(1.5)^{2}=\frac{1}{2}\times550\times2.25 = 618.75$ J.

Step3: Calculate the final kinetic energy

Given $m = 550$ kg and $v_f = 3.2$ m/s. Using the kinetic - energy formula $K_f=\frac{1}{2}mv_f^{2}=\frac{1}{2}\times550\times(3.2)^{2}=\frac{1}{2}\times550\times10.24 = 2816$ J.

Step4: Calculate the work done

$W=K_f - K_i=2816 - 618.75=2197.25\approx2197$ J.

Answer:

B. 2197 J