Q#13 (Past Exam Paper – June 2010 Paper 23 Q3)
A cyclist is moving up a slope that has a constant gradient. The cyclist takes 8.0 s to climb the slope.
The variation with time t of the speed v of the cyclist is shown in Fig. 3.1.
Fig. 3.1
(a) Use Fig. 3.1 to determine the total distance moved up the slope. [3]
(b) The bicycle and cyclist have a combined mass of 92 kg.
The vertical height through which the cyclist moves is 1.3 m.
(i) For the movement of the bicycle and cyclist between t = 0 and t = 8.0 s,
1. use Fig. 3.1 to calculate the change in kinetic energy, [2]
2. calculate the change in gravitational potential energy. [2]
(ii) The cyclist pedals continuously so that the useful power delivered to the bicycle
is 75 W.
Calculate the useful work done by the cyclist climbing up the slope. [2]
(c) Some energy is used in overcoming frictional forces.
(i) Use your answers in (b) to show that the total energy converted in overcoming
frictional forces is approximately 670 J. [1]
(ii) Determine the average magnitude of the frictional forces. [1]
(d) Suggest why the magnitude of the total resistive force would not be constant. [2]
Solution:
(a)
{Distance travelled is given by the area under the graph.}
evidence of use of area below the line
distance = 39 m (allow ±0.5 m)
(b)
(i)
1.
EK = ½ mv2
{At time t = 0 s, speed = 6 m/s and at time t = 8 s, speed = 3 m/s}
Δ EK = ½ × 92 × (62 – 32)
Δ EK = 1240 J
2.
EP = mgh
{Vertical height = 1.3 m}
Δ EP = 92 × 9.8 × 1.3
Δ EP = 1170 J
(ii)
{Power = Energy / time
Energy = Power × time}
E = Pt
E = 75 × 8
E = 600 J
(c)
(i)
{Initially, the cyclist had kinetic energy. The kinetic energy decreases with time as the speed decreases.
Change in KE = 1240 J (as calculated in (b)(i)1.)
The cyclist pedals and does work. So, in addition to the KE the cyclist also has energy from pedaling.
Work done = 600 J (as calculated in (b)(ii))
As the cyclist rises, some of its energy is converted in GPE.
Change in GPE = 1170 J (as calculated in (b)(i)2.)
Finally, some energy is used to overcome frictional forces. Let this energy be E.}
{From the conservation of energy.,
Energy possessed by cyclist = Energy converted into other forms
Change in KE + Work done = Change in GPE + E
1240 + 600 = 1170 + E}
energy E = (1240 + 600) – 1170
energy E = 670 J
(ii)
{Work done against friction = Frictional force × distance moved (along slope)}
force = 670 / 39 = 17 N
(d)
Frictional forces include air resistance.
Air resistance decreases with decrease of speed.
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