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WEshruth · Answer 1 · 2023-11-20T22:12:11+0000

We will have the following:

First, we convert the velocities to m/s, that is:

$\begin{cases}v_1=30(km)/(h)\cdot\frac{1000m}{1\operatorname{km}}\cdot(1h)/(60m)\cdot(1m)/(60s)\Rightarrow v_1=(25)/(3)m/s \\ \\ v_2=60\frac{\operatorname{km}}{h}\cdot\frac{1000m}{1\operatorname{km}}\cdot(1h)/(60m)\cdot(1m)/(60s)\Rightarrow v_2=(50)/(3)m/s\end{cases}_{}$

Then we will have the follwing:

$W=(1)/(2)mv^2_f-(1)/(2)mv^2_i\Rightarrow W=(1)/(2)(1200\operatorname{kg})((50)/(3)m/s)^2-(1)/(2)(1200\operatorname{kg})((25)/(3)m/s)^2$
$\Rightarrow W=125000\operatorname{kg}\cdot m^2/s^2\Rightarrow W=125000J$

So, the work done was of 125000 Joules.

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