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Work, energy and power Questions
1. (a) State the law of conservation of energy
(b) The graph below shows the potential energy against displacements for a body of mass 80g
The body oscillates about point R. Calculate the velocity of the body at:
(i) P and T
(ii) Q and S
(iii) at R
(c) A wheel and axle are used to raise a load of 280N by a force 40N applied to the rim of the
wheel. If the radii of the rim and axle are 70cm and 5cm respectively, calculate:
(i) The mechanical advantage
(ii) The velocity ratio
(iii) The efficiency
2. (a) A bicycle has wheels 66 cm in diameter. Its crank wheel has 44 teeth and the rear sprocket
16 teeth. The crank radius is 16.5 cm.
(i) Determine the radius of the rear sprocket.
(ii) The bicycle moves when the rear sprocket is made to move. Hence determine the
velocity ratio.
(b) A man uses a block and tackle mechanism of velocity ratio 6 to lift a car engine
smoothly through a height of 1 m in 5s. The man applies a force of 300N while the
mass of the engine is 120 kg. Determine:
(i) The mechanical advantage of the pulley system.
(ii) its efficiency.
3. (a) Define work and state its S.I units
(b) A crane lifts a load 500kg through a vertical distance of 4m in 8 seconds. Determine:
(i) Work done by the crane
(ii) Power developed by the crane
(iii) Efficiency of the crane given that it is operated by an electric motor rated 2.8Kw
(iv) State two effects which contribute to the efficiency being less than 100%
4. A load of 100N is raised using the system in the figure below by an effort.
Given that the efficiency of the machine is 90%, calculate the minimum effort.
Work, energy and power Answers
1. (a) The law of conservation of energy states that the sum of kinetic energy and potential energy of a system is a constant
(b) (i) At P and T potential energy is a maximum and kinetic energy is a minimum. Hence
velocity is zero (2mks)
(ii) At Q and S P.E has reduced by 0.1J. This equals the K.E
K.E = ½ MV2
0.1 = ½ x 0.8V2
0.1 = 0.4V2
0.1 = V2
V2 = ¼ = 0.25
V = 0.5m/s
(iii) At R, auP.E has been converted to K.E velocity now is a maximum
So, 0.2 = ½ MV2
0.4 = V2
V = 0.4m/s
V = 0.64m/s
(c) (i) M.A = L = 280N = 7 (2mks)
E 40N
(ii) V.R = P = 70 = 14 (2mks)
R 5
(iii) n = M.A x 100%
V.R
= 7 x 100%
14
= 50% (2mks)
2. a) (i) CR = 2R = No of teeth draw
2r No. of teeth of driven 1
∴
16.5 cm = 44
r 16
r = 16.5 cm x 16 1
44
R = 6 cm 1
(ii) V.R. = R
1
r
= 16.5 cm
1
6 cm
= 2.75 1
b) (i) M.A. = L 120kg x 10N/kg = 1200N 1
E
= 1200N
1
300N
= 4 1
(ii) Its efficiency of:
D = M.A x 100%
1
V.R.
= 4 x 100%
6
= 66.67% 1
3. (a) Work is said to be done when the body on which a force is applied moves in the direction
of force S.I unit if the Joule, J or (Nm);
(b) (i) Work done = P.E gained
= mgh
= 500 x 4 x 10;
= 20,000J;
(ii) Power = work done
time taken
= 20,000
8
= 2.5KW or (2500 watts);
(iii) Efficiency = work output x100;
work input
= 2.5 x 100;
2.8
= 89.29%;
(iv) – Friction between movable parts
- Sound due to moving parts
- – heat –some of the electrical energy is converted to unnecessary heat
