Question #145599
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
1
Expert's answer
2020-11-23T10:28:19-0500

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


a=Δvta = \dfrac{\Delta v}{t}

where Δv=vfvi=30m/s20m/s=10m/s\Delta v =v_f-v_i = 30m/s - 20m/s = 10m/s is the velocity increase (difference between final vfv_f and initial viv_i velocities). Then the engin force will be (according to the Newton's second law):


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

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

The distance travelled in time tt is:


d=v0t+at22=v0t+Δvt2=v0t+Δvt2d = 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=FdA=mΔvt(v0t+Δvt2)=mΔv(v0+Δv2)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=35010(20+102)=87500 JA = 350\cdot 10\cdot \left( 20 + \dfrac{10}{2} \right) = 87500\space J

Answer. 87500 J.


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