A mass hangs on the end of a massless rope. The pendulum is held horizontal and

Question

A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.2 m/s and the tension in the rope is T = 19.7 N.

How long is the rope?
What is the mass?
3) If the maximum mass that can be used before the rope breaks is mmax = 1.43 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
4)Now a peg is placed 4/5 of the way down the pendulum

Explanation / Answer

since released from height = length of rope initial potential energy = mgl and total energy at bottom position = 1/2 m* v^2 no from energy conservation mgl=1/2 m*v^2 ====>2mg=mv^2/l so l= 2.2^2/(2*9.8)=0.247 m now tension of rope = mg+mv^2/l = 19.7 = 3*mg so m= 0.67kg max tension = 3*Mmax*g so max tension = 42.042N calculate 4th and 5th from conservation of energy

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