Jamal pulls a 250-N sled up a 29.0° slope at constant speed by a force of 200 N.Near the top of the hill he releases the sled.With what acceleration does the sled go down the hill? A)7.84 m/s² B)9.30 m/s² C)3.32 m/s² D)1.66 m/s² E)2.66 m/s²

Question

Quizplus · Accepted Answer

Since the sled is moving up the slope at a constant speed, the net force acting on it must be zero. Therefore, the force of friction acting on the sled must be equal and opposite to the force applied by Jamal. We can find the force of friction using:

Force of friction = Force applied - Component of weight parallel to slope
Force of friction = 200 N - (250 N * sin 29°)
Force of friction = 39.2 N

When Jamal releases the sled, the only force acting on it is its weight (250 N) down the slope. Since there are no other forces to oppose its motion, the sled will accelerate down the slope. We can use Newton's second law to find the acceleration:

Acceleration = net force / mass
Acceleration = (250 N - 39.2 N) / (250 kg * sin 29°)
Acceleration = 1.66 m/s^2

Jamal Pulls a 250-N Sled Up a 29