A rocket is attached to a crate with a combined mass of 1330 kg. The crate is launched with 242 N of thrust down a frictionless icy hill sloped at 16.4° from the horizontal. What is the magnitude of acceleration of the crate while it travels down the hill?
Question
A rocket is attached to a crate with a combined mass of 1330 kg. The crate is launched with 242 N of thrust down a frictionless icy hill sloped at 16.4° from the horizontal. What is the magnitude of acceleration of the crate while it travels down the hill?
Solution
Sure, let's solve this step by step.
Step 1: Calculate the force of gravity acting on the crate. This is given by the formula F = m*g, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).
F = 1330 kg * 9.8 m/s^2 = 13034 N
Step 2: Calculate the component of the gravitational force that acts parallel to the slope. This is given by the formula F_parallel = F * sin(θ), where θ is the angle of the slope.
F_parallel = 13034 N * sin(16.4°) = 3700.6 N
Step 3: Calculate the net force acting on the crate. This is the difference between the thrust of the rocket and the component of the gravitational force parallel to the slope.
F_net = Thrust - F_parallel = 242 N - 3700.6 N = -3458.6 N
The negative sign indicates that the net force is acting against the direction of the rocket's thrust.
Step 4: Calculate the acceleration of the crate. This is given by the formula a = F_net / m.
a = -3458.6 N / 1330 kg = -2.6 m/s^2
So, the magnitude of the acceleration of the crate while it travels down the hill is 2.6 m/s^2.
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