# Questions for Work, Energy And Power (JC) Set 1

a) i) Define power

ii) Hence, derive the equation P = Fv, where P is power, F is force and v is speed.

b) A motorized bicycle has a maximum of 0.23 hp, horse power. 1 hp = 746 W. i) Along a horizontal road AB, the cyclist travels at a constant maximum speed of 33 km h-1. Calculate the total frictional force exerted on the system (bicycle and cyclist), if the efficiency of the motor is 60%.

ii) The cyclist then travels along a road BC inclined at 10° to the horizontal. Is it possible for the cyclist to move up the slope at the same constant speed of 33 km/h? Explain your answer.

iii) If the cyclist is to travel along the road inclined at 10° at the same constant speed of 33 km/h, calculate the new power required in hp(horse power) of the motor. The total frictional force acting on the system is now 0.8 times that in b)i), and the total mass of the system is 90 kg.

a) i) Power is defined as the rate of doing work.ii) P = dW/dt

P = d(Fx)/dt, where F is a constant force acting on a body and x is the displacement along the direction of the force.

Since force is constant,
P = F (dx/dt)
P = Fv

b) i) Since P = Fv,
60/100 x 0.23 x 746 = F x (33 x 103/3600)
F = 11.231 N

Since speed is constant, the forward driving force F is equal to the total frictional force f acting on the system. Hence, f = 11.2 N.

b) ii) No, it is not possible. This is because part of the power has to be used to do work against gravitational force.

b) iii) P = (f + mgsin 10°)v
P = [(0.8 x 11.231) + (90 x 9.81 x sin 10°)] x (33 x 103/3600)
P = 1487.7 W
P = 1487.7 / 746 Hp
P = 2.0 hp

Back To Work, Energy And Power ### 4 thoughts on “Questions for Work, Energy And Power (JC) Set 1”

1. In part biii) shouldn’t we still incorporate the efficiency of the motor and divide by 0.6 at the end to get 3.38 hp?

2. How do you get mgsin10?

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