Question

A 3.00-kg mass rests on the ground. It is attached to a string which goes vertically to and over an ideal pulley. A second mass is attached to the other end of the string and released. The 3.00-kg mass rises 50.0 cm in 1.00 s. How large was the other mass?

Answer 1

Answer:3.68 kg

Step-by-step explanation:

Given

mass
`m_1=3 kg`

distance traveled by
`m_1` is 50 cm in 1 s

using
`s=ut+(at^2)/(2)`

`0.50=0+(a\cdot 1^2)/(2)`

`a=1 m/s^2`

Suppose T is the tension in string and
`m_2` is the mass of other body

For
`m_1`

`T-m_1g=m_1a`

`T=m_1(g+a)`

for other body

`m_2g-T=m_2a`

`T=m_2(g-a)`

Equating value of Tension

`m_2=m_1* (g+a)/(g-a)`

`m_2=3* (10.8)/(8.8)`

`m_2=3.68 kg`