A lvl H2 Phy: Work Energy Power

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 E_k = 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(v_i + v_f) t
and acceleration = (v_f - v_i) / t

W = m * (v_f - v_i) / t * ½(v_i + v_f) t
W = ½ m(v_f - v_i)(v_i + v_f)
W = ½ m v_f² - ½ m v_i²

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 E_k = 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.

A lvl H2 Phy: Work Energy Power

In a children's game, small balls are thrown at wood blocks in order to turn them over. One such block, of mass 150g with each side of length 10cm, is shown in the figure below.

In order to turn the block over, the centre of gravity C of the block must raised so that C is vertically above the corner A as shown below.

(i) For the block as shown in the above figure
(1) Calculate the vertical heigh through which the centre of gravity has been raised.
(2) show that the gain in potential energy of the block is approximately 0.031 J


(ii) The block is struck by a ball of mass 11 g travelling horizontally towards C, as shown below.

The collision is perfectly elastic and, without sliding, the block turns about the corner A. The block is just able to reach the position in the 2nd figure (with C directly above A). 25% of the kinetic energy of the ball is transferred to the block. Calculate
(1) the kinetic energy of the ball just before it strikes the block,
(2) the speed with which the ball strikes the block,
(3) the speed with which the ball rebounds from the block.

(iii) For the collision in (ii) of the ball with the block, calculate
(1) the change in momentum of the ball,
(2) the average force on the block, assuming the ball and the block are in contact for 0.15 s.

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Answer:

(i)
(1) Distance between C and A = ½ √(10² + 10²) = 7.07 cm
=> vertical height in first diagram is 5 cm
=> vertical height in 2nd diagram is 7.07 cm
the centre of gravity has been raised by 7.07 cm -5 cm = 2.07 cm

(2) Gain in G.P.E. = mgh
= (150 * 10^-3) (9.81) (2.07 / 100)
= 0.031 J (2 s.f.)


(ii)
(1) Let K = the kinetic energy of the ball just before it strikes the block. Since 25% of K is transferred to the block
=> 0.25 K = Gain in potential energy of the block
=> 0.25K = 0.031 => K = 0.124 J

(2) Let u = speed with which ball strikes the block.
K = ½ mu² => u = √(2K/m) = √(2 * 0.124 / (11 * 10^-3))
u = 4.75 ms^-1

(3) After the collision, kinetic energy of ball = 0.75K
Let v = speed with which ball rebounds from block
0.75K = ½ mv² => v = √(2*0.75K/m)
v = = √(2 * 0.75 * 0.124 / (11 * 10^-3))
v = 4.11 ms^-1


(iii)
(1) change in momentum of ball = mv - mu
= m (v - u)
= (11 * 10^-3) (4.11 - 4.75)
= -7.04 * 10^-3 kg ms^-1

(2) By Newton's 2nd Law,
Force on the ball = Rate of change of momentum of ball

By Netwon's 3rd law,
Force on the block = force on the ball (numerically)
= rate of change of momentum of ball
= -7.04 * 10^-3 / 0.15
= 0.047 N

A lvl H2 Phy: Work Energy Power

A 1000kg car climbs a hill with 4 degree inclined at a constant speed of 12m/s.
If the mechanical power output of the engine is 20kw, find the force of air resistance acting on the car. Ignore the friction between the car tyres and the road

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Answer:

P=FxV

so F=P/V
=20000/12
=1.6 x 10³ N

A lvl H2 Phy: Work Energy Power

A 20 cm long spring stretches to a length of 22cm when pulled horizontally with a force of 100N. It is placed vertically on the ground and 10.2 kg block is held 10cm above the spring. The block is dropped, hits the spring and compresses it. What is the height of the spring at the point of maximum compression?

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Answer:

F = ke, F = 100N, e = 0.02m
k = F/e = 5000N/m

compression gives stored energy
stored energy = 1/2 k e²

loss in PE = mg(0.1 + e) = gained in stored energy = 1/2 k e²
100(0.1 + e) = 2500 e²
1 + 10e = 250 e²

using formula,
e = 0.0863 or -0.0463 (reject)

compression of 8.63 cm
Height of spring = 20 - 8.63 = 11.4 cm