Question

A 8.9 kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 2.9 kg mass. Use conservation of energy to determine the final speed of the masses after the heavier mass has fallen (starting from rest) 7.7 m .The acceleration of gravity is 9.8 m/s² . Answer in units of m/s.

Answer 1

Answer:

The final speed of the masses is 8.75 m/s.

Explanation:

Given that,

Mass = 8.9 kg

Other mass = 2.9 kg

Distance = 7.7 m

Let mass $M_{s}$ start at its 0 Potential energy position.

Let mass $M_{l}$ start with potential energy from its position 4.6 m over final position.

We need to calculate the final speed of the masses

Using conservation of energy

$K.E_{s}+P.E_{s}+K.E_{l}+P.E_{l}=K.E'_{s}+P.E'_{s}+K.E'_{l}+P.E'_{l}$

$\dfrac{1}{2}m_{s}v^2+m_{s}gh+\dfrac{1}{2}m_{l}v^2+m_{l}gh=\dfrac{1}{2}m_{s}v^2+m_{s}gh+\dfrac{1}{2}m_{l}v^2+m_{l}gh$

Put the value in the equation

$0+0+0+8.9\times9.8\times7.7=\dfrac{1}{2}m_{s}v^2+2.9\times 9.8\times7.7+\dfrac{1}{2}m_{l}v^2+0$

$8.9\times9.8\times7.7-2.9\times9.8\times7.7=\dfrac{1}{2}v^2(8.9+2.9)$

$\dfrac{1}{2}v^2\times11.8=452.76$

$v^2=\dfrac{452.76\times2}{11.8}$

$v=8.75\ m/s$

Hence, The final speed of the masses is 8.75 m/s.