Question 13
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]
Reference: Past Exam Paper – June 2010 Paper 23 Q3
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.
How do we take out the area when there is curved line, my area is coming 36m, kindly guide..
ReplyDeleteFINALLY SOMEONE ASKED IT, HOW DO WE CALCULATE THE AREA UNDER LINE, WHICH IS CURVED?
ReplyDeletewe do it by estimation.
DeleteLook at the area under the curve and try to break it down into triangles, rectangles, trapezium ... and calculate the area of each piecewise. then take the total area.
for example, you could break in into 2 trapezia: one from t = 0 to t = 4 and the other from t = 4 to t = 8.
of course, if you look carefully, the area obtained for the first trapezium would be a little less as we have not accounted for a small part just above the trapezium. we can count the number of small squares and add it.
on both axes,
10 small squares represent 2 (s or m/s)
so 1 small square represents 0.2
area of 1 small square = 0.2 x 0.2 = 0.04 m
take the additional number of small squares times 0.04m.
then add all.