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d) 0.24 m e) 0.28 m 016: A 4 kg mass is projected down a rough ( )circular track (R-2 m) as
shown. The speed of the mass at point A is 4 m/s, and at point B, it is 6 m/s. How much work is
done on the mass between A and B by the force of friction? a)-19.2 b)-38.4 c) -28.8 d)-57.6 e)-
9.4 017:An elevator has a weight of 1600 nn
Solution
16.
Result : (b)
Explanation:
By conservation of energy principle :
K A +U A -W f = K B
where , K A & K B are kinetic energies at A and B
W f is the work done by friction force
U A is the potential energy at A
i.e. W f = K A +U A -K B
= (1/2)mv A
2
+ mgR - (1/2)mv B
2
= (1/2)*(4)*(4) 2
+ (4)*(9.81)*(2) - (1/2)*(4)*(6) 2
= 32 J + 78.48 J - 72 J
= 38.4 J
As it acts in the direction opposite to that of motion , hence it has negative sign.
So work done by friction force is : W f = -38.4 J
d) 0-24 m e) 0-28 m 016- A 4 kg mass is projected down a rough ( )circ.docx

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d) 0-24 m e) 0-28 m 016- A 4 kg mass is projected down a rough ( )circ.docx

  • 1. d) 0.24 m e) 0.28 m 016: A 4 kg mass is projected down a rough ( )circular track (R-2 m) as shown. The speed of the mass at point A is 4 m/s, and at point B, it is 6 m/s. How much work is done on the mass between A and B by the force of friction? a)-19.2 b)-38.4 c) -28.8 d)-57.6 e)- 9.4 017:An elevator has a weight of 1600 nn Solution 16. Result : (b) Explanation: By conservation of energy principle : K A +U A -W f = K B where , K A & K B are kinetic energies at A and B W f is the work done by friction force U A is the potential energy at A i.e. W f = K A +U A -K B = (1/2)mv A 2 + mgR - (1/2)mv B 2 = (1/2)*(4)*(4) 2 + (4)*(9.81)*(2) - (1/2)*(4)*(6) 2 = 32 J + 78.48 J - 72 J = 38.4 J As it acts in the direction opposite to that of motion , hence it has negative sign. So work done by friction force is : W f = -38.4 J