Question

Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope with a force of 250 N. The other has a mass of 60 kg. What is the acceleration of each man?

Answer 1

The acceleration of man 1 and 2 is
`3.125\ m/s^2` and
`4.167\ m/s^2`.

Step-by-step explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

`a_1=(F)/(m_1)\\\\a_1=(250)/(80)\\\\a_1=3.125\ m/s^2`

Let a₂ is acceleration of man 1,

F = m₂a₂

`a_2=(F)/(m_2)\\\\a_2=(250)/(60)\\\\a_2=4.167\ m/s^2`

So, the acceleration of man 1 and 2 is
`3.125\ m/s^2` and
`4.167\ m/s^2`.