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 = 2.3 m/s and the tension in the rope is T = 22.4 N.

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 = 2.3 m/s and the tension in the rope is T = 22.4 N.
1. What is the mass?
2. If the maximum mass that can be used before the rope breaks is mmax = 1.72 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)

Joe · Accepted Answer

Second equation is correct, First equation is (1/2mv^2)/mg => (1/2v^2)/g

Anonymous · Answer

1.What is the mass find the length of the rope v^2/(2*g)=L find the mass (L*T)/(v^2+L*g)=m

Kay · Answer

(1/2v^2)/g is the same as v^2/(2g) so both work!