Question

A rocket is attached to a person in a sled with a combined mass of 48 kg. The sled is launched with 431 N of thrust up a frictionless icy hill sloped at 16.2° from the horizontal. What is the magnitude of acceleration of the sled while it travels up the hill?

0.821 m/s2.
2.73 m/s2.
11.7 m/s2.
6.25 m/s2.

Answer 1

Step-by-step explanation:

To determine the acceleration of the sled, we need to balance the net force acting on it with its weight.

The weight of the sled and person is given by:

Weight (W) = mass (m) x acceleration due to gravity (g) = 48 kg x 9.8 m/s^2 = 470.4 N

The weight acts vertically downward, perpendicular to the slope of the hill. The net force acting on the sled can be found by subtracting the force of friction from the thrust:

Net force (F) = thrust (T) - weight (W) x sin(θ)

where θ is the angle between the horizontal and the slope of the hill, given as 16.2°.

The force of friction is zero in this case, as the sled is moving up a frictionless icy hill.

So, the net force is simply equal to the thrust:

Net force (F) = thrust (T) = 431 N

The acceleration of the sled is given by:

acceleration (a) = net force (F) / mass (m) = 431 N / 48 kg = 9 m/s^2

Therefore, the magnitude of the acceleration of the sled while it travels up the hill is 9 m/s^2.