A41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible mass. the girl exerts a horizontal 4.0 n force on the rope. what are the acceleration magnitudes of (a) the sled and (b) the girl? (c) how far from the girl's initial position do they meet?

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

where is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

where is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

no sign of dia of fuel cell

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.