Q4.6 Work and kinetic energy II (S)
A truck pulling a trailer of mass 4000 kg accelerates steadily from rest to a speed of 72 kmh-1 in 40 s.- How far does the truck travel during the period of acceleration?
- Calculate the work done by the truck on the trailer over this distance and show that it is equal to the increase in kinetic energy of the trailer.
Solution
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Solution
From the magnitude of the acceleration, we can find the force applied to the trailer by the truck, and the distance travelled.
First calculate the acceleration: 72 kmh-1=72×1000/3600=20 ms-1. Thus, the acceleration is
a=20/40=0.5 ms-2
- The truck travels a distance
![\[ x = \frac{1}{2} a t^2 = 400\,m \]](mastermathpng-0.png)
The magnitude of the (constant) force applied by the truck on the trailer is
F=ma=2000 Ntherefore the work done by this force over 400m is
W=2000×400=8×105 JThe final kinetic energy of the trailer is
![\[ K = \frac{1}{2} m v^2 = 2000 \times 20 \times 20 = 8 \times 10^5\,J \]](mastermathpng-1.png)
Not entirely unexpected!