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

Answer:

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

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

$a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2$

Let a₂ is acceleration of man 1,

F = m₂a₂

$a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2$

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