Question

Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the man is 95 kg, and the mass of the woman is 51 kg. The woman pushes on the man with a force of 60 N due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman

Answer 1

Answer:

a₁ = 0.63 m/s²  (East)

a₂ = -1.18 m/s²  (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a   (⇒)

F = m₁*a₁         ⇒      60 N = 95 Kg*a₁    

⇒  a₁ = (60N / 95Kg) = 0.63 m/s²  (⇒)  East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a   (⇒)

F = -m₂*a₂         ⇒      60 N = -51 Kg*a₂    

⇒  a₂ = (60N / 51Kg) = -1.18 m/s²  (West)

For every case we apply Newton’s 3 d Law