November 1999 P3 Q1
(a) Starting with the definition of work, deduce the change in the gravitational potential energy of a mass m, when moved a distance h upwards against a gravitational field of field strength g.
(b) By using the equations of motion, show that the kinetic energy Ek of an object of mass m travelling with speed v is given by Ek = ½mv²
(c) A cyclist, together with his bicycle, has a total mass of 90 kg and is travelling with a constant speed of 15 ms-1 on a flat road at A, as illustrated in the figure below. He then slide down a small slope to B so descending 4.0m.
Calculate
(i) the kinetic energy at A,
(ii) the loss of potential energy between A and B, and
(iii) the speed at B, assuming that all the lost potential energy is transformed into kinetic energy of the cyclist and the bicycle.
(d)
(i) A cyclist is travelling at a constant speed of 15 ms-1 on a level road provides a power of 240 W. Calculate the total resistive force.
(ii) The cyclist now travels at a higher constant speed. Explain why the cyclist needs to provide a greater power.
(e) It is often stated that many forms of transport transform chemical energy into kinetic energy. Explain why a cyclist travelling at constant speed is not making this transformation. Explain what transformations of energy are taking place.
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Answer:
(a) Work done is equal to the force times the distance moved along the direction of the force.
Increase in GPE of mass m
= work done against the gravitational field strength of g in raising the mass a distance of h upwards
= gravitational force on the mass m times distance moved (upwards)
= mg * h
= mgh (deduced)
(b) Work done by a force F in displacing a mass m by displacement s is given by
W = F s = (ma) s
Since displacement, s = ½(vi + vf) t
and acceleration = (vf - vi) / t
W = m * (vf - vi) / t * ½(vi + vf) t
W = ½ m(vf - vi)(vi + vf)
W = ½ m vf² - ½ m vi²
The above is the change in kinetic energy, or work done by a force F in displacing a mass m by displacement s. This shows that the kinetic energy is Ek = ½mv²
(c)
(i) K.E. at A = ½mv²
= ½(90)(15)²
= 1.01 * 104 J.
(ii) Loss in PE = mgh
= 90 * 9.81 * 4.0
= 3.53 * 103 J.
(iii)
Let the speed at B be v'
Gain in kinetic energy = loss in potential energy
½mv'² - 1.01 * 104 = 3.53 * 103
½mv'² = 1.01 * 104 + 3.53 * 103
v' = 17.4 ms-1
(d)
(i) Power = Fv
F = 240 / 15 = 16N
(ii) Total resistive force is proportional to the speed of the cyclist. Since the cyclist is now travelling at a higher constant speed, the total resistive force is greater. Thus, a greater power is required to maintain a higher constant speed.
(e) When chemical energy is transformed into kinetic energy, there will be an increase in kinetic energy. There will thus be an increase in speed. Since the cyclist travels at a constant speed, there's obviously no change in kinetic energy. Thus, there's no transformation from chemical energy to kinetic energy.
The chemical energy is transformed into the rotational kinetic energy of the bicycle wheels as well as heat energy as the cyclist moves against the resistive forces.
