Answer to Question #145599 in Physics for Mya

Question

A 350kg motorcycle is moving at 20m/s. It then speeds up to 30m/s. How much work did the engine do to accelerate? *
5 points

Answer 1

Let $t$ be the time during which motorcycle was being accelerated. Then, the acceleration is:

$$a = \dfrac{\Delta v}{t}$$

where $\Delta v =v_f-v_i = 30m/s - 20m/s = 10m/s$ is the velocity increase (difference between final $v_f$ and initial $v_i$ velocities). Then the engin force will be (according to the Newton's second law):

$$F = ma = m\dfrac{\Delta v}{t}$$

where $m = 350kg$ is the mass of the motorcycle.

The distance travelled in time $t$ is:

$$d = v_0t + \dfrac{at^2}{2} = v_0t + \dfrac{\Delta vt}{2} = v_0t + \dfrac{\Delta vt}{2}$$

The work done by the engin is:

$$A = Fd\\ A = m\dfrac{\Delta v}{t}\left( v_0t + \dfrac{\Delta vt}{2} \right) = m\Delta v\left( v_0 + \dfrac{\Delta v}{2} \right)$$

Substituting the numbers, obtain:

$$A = 350\cdot 10\cdot \left( 20 + \dfrac{10}{2} \right) = 87500\space J$$

Answer. 87500 J.