A car of mass 700 kg is travelling along a straight horizontal road. The resistance to motion is constant and equal to 600 N.
- Find the driving force of the car’s engine at an instant when the acceleration is 2 m s-2.
- Given that the car’s speed at this instant is 15 m s-1, find the rate at which the car’s engine is working.
Solution
(i) Using Newton's second law, the net force on the car is given by:
\(F = ma\)
where \(m = 700 \text{ kg}\) and \(a = 2 \text{ m s}^{-2}\).
The driving force \(DF\) minus the resistance force equals the net force:
\(DF - 600 = 700 \times 2\)
\(DF - 600 = 1400\)
\(DF = 1400 + 600 = 2000 \text{ N}\)
(ii) The power \(P\) is given by:
\(P = Fv\)
where \(F = 2000 \text{ N}\) and \(v = 15 \text{ m s}^{-1}\).
\(P = 2000 \times 15 = 30000 \text{ W}\)
Thus, the rate of working is 30000 W or 30 kW.
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