RI Sec 3 EOY 1998 P2 Q4
A student releases a trolley of mass 0.20 kg from the top of an inclined runway of height 0.40 m as shown in the figure below.
The speed of the trolley at the bottom of the runway is 1.0 m/s
(a) Calculate
(i) the potential energy of the trolley at the top of the runway,
(ii) the kinetic energy of the trolley at the bottom of the runway,
(iii) the work done by the trolley against friction.
The trolley experiences a constant frictional force of 0.50 N throughout its motion, and takes 2.8 s to reach the bottom of the runway.
(b) Calculate
(i) the acceleration of the trolley,
(ii) the net force pulling the trolley down the runway,
(iii) the force of gravity pulling the trolley down the runway.
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Answer:
(a)
(i) Potential energy = mgh = 0.20 * 10 * 0.40 = 0.80 J
(ii) kinetic energy of trolley at bottom = ½ mv²
= ½ (0.20)(1.0)²
= 0.10 J
(iii) work done aganist friction = 0.80 J - 0.10 J = 0.70 J
(b)
(i) Acceleration of trolley = (1.0 m/s) / (2.8 s)
= 0.36 m/s²
(ii) Using F = ma,
net force = 0.20 * 0.36
= 0.072 N
(iii) Net force downwards = force of gravity down runway - frictional force
Force of gravity on trolley down runway = net force + frictional force
= 0.072 N + 0.50 N
= 0.57 N (2 s.f.)
