A 90kg woman and a 60kg boy are standing at rest on a frictionless frozen lake. The boy pushes the woman with a 40N horizontal force. Calculate the acceleration of the woman


3600 m/s2



0.66 m/s2



26.7 m/s2



0.44 m/s2



30 m/s2



1.5 m/s2

What happens to them?

The woman will move slowly, and the boy will move faster in the opposite direction.


The woman and the boy will move together with equal speeds.


The woman will move away, but the boy will remain at rest.


The boy will move away, but the woman will remain at rest.


The woman will move slowly, and the boy will move faster in the same direction.


The woman and the boy will move apart at equal speeds.

Respuesta :

Answer:

The acceleration of the woman is 0.44 m/s²

Explanation:

Given;

mass of the woman, m₁ = 90 kg

mass of the boy, m₂ = 60 kg

The force applied by the boy, f₂ = 40 N

The net horizontal force on the woman = 40 N

Apply Newton's second law of motion to determine the acceleration of the woman;

f = ma

a = f / m

a = 40 / 90

a = 0.44 m/s²

Therefore, the acceleration of the woman is 0.44 m/s²