An empty supermarket trolley of mass 10 kg is moving with a speed of 8 m s-1. An identical trolley is loaded with 10 kg of goods and moves with a speed of 4 m s-1. Both trolleys are brought to rest over the same stopping distance.
(a) Calculate the kinetic energy of the empty trolley.
(b) Calculate the kinetic energy of the loaded trolley.
(c) Compare the braking force required to stop the empty trolley and the loaded trolley.
(d) Name the principle applied in solving part (c).
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Given,
- Mass of the empty trolley (m1) = 10 kg
- Speed of the empty trolley (v1) = 8 m s-1
- Mass of the loaded trolley (m2) = 10 + 10 = 20 kg
- Speed of the loaded trolley (v2) = 4 m s-1
(a) Kinetic energy of the empty trolley
$\text K_1 = \frac{1}{2}\text m_1 {\text v_1}^2 = \frac{1}{2} \times 10 \times 8 \times 8 = 320 \text { J}$
Hence, the kinetic energy of the empty trolley is 320 J.
(b) Kinetic energy of the loaded trolley
$\text K_2 = \frac{1}{2}\text m_2 {\text v_2}^2 = \frac{1}{2} \times 20 \times 4 \times 4 = 160 \text { J}$
Hence, the kinetic energy of the loaded trolley is 160 J.
(c) The work done by the braking force in stopping a trolley is equal to its kinetic energy. If the stopping distance is s and the braking force is F, then
F × s = kinetic energy
So,
$\frac{\text F_1}{\text F_2} = \frac{\text K_1 / \text s}{\text K_2 / \text s} = \frac{\text K_1}{\text K_2} = \frac{320}{160} = \frac{2}{1}$
Hence, F1 : F2 = 2 : 1, i.e., the braking force needed for the empty trolley is twice that needed for the loaded trolley.
(d) The work-energy theorem, i.e., the work done by the braking force is equal to the change in the kinetic energy of the trolley.