Question

A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?

Answer 1

Answer:

$10\: \mathrm{J}$

Explanation:

The kinetic energy of an object is $KE=\frac{1}{2}mv^2$, where $m$ is the mass of the object and $v$ is the velocity of the object.

The toy car's initial kinetic energy is $KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}$.

After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.

To find the final velocity of the toy car, we can use kinematic equation $v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}$

We are given $v_i=5\: \mathrm{m/s}$ and $\Delta x = 2\: \mathrm{m}$.

To find acceleration:

$F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}$.

Now substitute $v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}$ into $v_f=\sqrt{v_i^2+2a\Delta x}$ to get $v_f\approx 5.92\: \mathrm{m/s}$.

Using this, we can find the final kinetic energy of the toy car is $KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2$.

Thus, the change in kinetic energy is $KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{ 10\: \mathrm{J} }$ (one significant figure).