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.
  1. How far does the truck travel during the period of acceleration?
  2. 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.
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Hint

Review S3.2 for the definition of work.
 

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

  1. The truck travels a distance
    \[ x = \frac{1}{2} a t^2 = 400\,m \]
  2. The magnitude of the (constant) force applied by the truck on the trailer is

    F=ma=2000 N

    therefore the work done by this force over 400m is

    W=2000×400=8×105 J

    The final kinetic energy of the trailer is

    \[ K = \frac{1}{2} m v^2 = 2000 \times 20 \times 20 = 8 \times 10^5\,J \]

    Not entirely unexpected!