Question

1. how much work is needed to slow down a 1200 kg vehicle from 80 km/h to 50 km/h?
2. how much work is needed to accelerate a 1.0 g insect from rest up to 12 m/s?

Explanation:

Step1: Recall the work - energy theorem

The work - energy theorem states that $W=\Delta K = K_f - K_i$, where $K=\frac{1}{2}mv^{2}$, $m$ is the mass and $v$ is the velocity.

Step2: For the insect problem

First, convert the mass of the insect to SI units. $m = 1.0\ g=0.001\ kg$, $v_i = 0\ m/s$, $v_f = 12\ m/s$.
$K_i=\frac{1}{2}mv_i^{2}=0$ (since $v_i = 0$).
$K_f=\frac{1}{2}mv_f^{2}=\frac{1}{2}\times0.001\ kg\times(12\ m/s)^{2}=\frac{1}{2}\times0.001\times144\ J = 0.072\ J$.
So, $W=\Delta K=K_f - K_i=0.072\ J$.

Step3: For the vehicle problem

Convert the velocities to SI units. $v_i = 80\ km/h=\frac{80\times1000}{3600}\ m/s=\frac{200}{9}\ m/s\approx22.22\ m/s$, $v_f = 50\ km/h=\frac{50\times1000}{3600}\ m/s=\frac{125}{9}\ m/s\approx13.89\ m/s$, $m = 1200\ kg$.
$K_i=\frac{1}{2}mv_i^{2}=\frac{1}{2}\times1200\ kg\times(\frac{200}{9}\ m/s)^{2}=\frac{1200}{2}\times\frac{40000}{81}\ J=\frac{24000000}{81}\ J\approx296296.3\ J$.
$K_f=\frac{1}{2}mv_f^{2}=\frac{1}{2}\times1200\ kg\times(\frac{125}{9}\ m/s)^{2}=\frac{1200}{2}\times\frac{15625}{81}\ J=\frac{9375000}{81}\ J\approx115740.7\ J$.
$W=\Delta K=K_f - K_i=\frac{9375000}{81}-\frac{24000000}{81}=-\frac{14625000}{81}\ J\approx - 180555.6\ J$. The negative sign indicates that work is done on the vehicle to slow it down.

Answer:

For the insect, the work needed is $0.072\ J$. For the vehicle, the work needed is approximately $- 180555.6\ J$.