Question

A 20-kg rocket sled is sliding up a 30 degree, 10-m high snow ski jumping hill from rest with constant thrust. The kinetic friction coefficient between the sled and snow is 0.2. The engine is turned off when the sled leaves the hill and it lands 50-m away from where it left the hill at ground level. The drag due to air resistance is negligible. What is the magnitude of thrust of the rocket sled?

Edit: This is the question in verbatim. this is all the information I have to go on.

Answer 1

A frictional force is given by, F_f = _k F_N = _k mg cos

F_f = (0.2) (20 kg) (9.8 m/s²) cos 30⁰

F_f = 33.9 N

We know that, K.E = W_friction

(1/2) m v² = F_f d

v² = 2 (33.9 N) (10 m)

v = 678 m²/s²

v = 26.03 m/s

Using equation of motion 3, we have

v² = v₀² + 2 a s

where, v₀ = 0 m/s

(26.03 m/s)² = 2 a (50 m)

a = 6.7 m/s²

Therefore, magnitude of thrust of the rocket sled will be given as :

F_net = F_thrust - F_f

m a = F_thrust - (33.9 N)

F_thrust = (20 kg) (6.7 m/s²) + (33.9 N)

F_thrust = 167.9 N