A 43 kg girl and a 7.1 kg sled are on the frictionless ice of a frozen lake, 12 m apart but connected by a rope of negligible mass. The girl exerts a horizontal 6.6 N force on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl

(a) Girl acceleration = 0.15 m/s², (b) Sled acceleration = 0.93 m/s²

Explanation:

Given:

M1=Mass of girl = 43 Kg

M2 = mass of Sled = 7.1 Kg

Force F = 6.6 N

To Find (a) a1 = girl acceleration (b) a2 = sled acceleration

Solution:

F1 = F2 = F (Newton's 3rd Law)

F=ma (Newton's 2nd Law)

(a) F = M1a1

⇒a1 = F/M1 = 6.6 N / 43 Kg = 0.15 m/s²

(b) a2 = F / M2 = 6.6 N / 7.1 Kg = 0.93 m/s²