A rocket is attached to a person in a sled with a combined mass of 55 kg. The sled is launched with 360 N of thrust up a frictionless icy hill sloped at 19° from the horizontal. What is the magnitude of acceleration of the sled while it travels up the hill?
weight = 55 g = 55*9.81 = 540 Newyons
component of weight down slide = 540 sin 19 = 176 N
so net force up slope = 360 - 176 = 184 N
a = 184/55 m/s^2
To find the acceleration of the sled while it travels up the hill, we can use Newton's second law of motion:
F = ma
Where F is the net force acting on the object, m is the mass of the object, and a is the acceleration.
In this case, the net force acting on the sled is the thrust provided by the rocket, which is 360 N. The mass of the sled and the person combined is 55 kg.
Now, we need to decompose the forces acting on the sled into horizontal and vertical components. Since the hill is sloped at an angle of 19° from the horizontal, we need to find the component of the force that acts parallel to the slope and the component that acts perpendicular (normal) to the slope.
The vertical component of the force is given by F_parallel = F * sin(theta), where theta is the angle of the slope.
F_parallel = 360 N * sin(19°)
F_parallel ≈ 120.26 N
The horizontal component of the force is given by F_perpendicular = F * cos(theta).
F_perpendicular = 360 N * cos(19°)
F_perpendicular ≈ 337.113 N
Now we can calculate the acceleration by using the net force, which is the horizontal component of the force.
a = F_perpendicular / m
a = 337.113 N / 55 kg
a ≈ 6.13 m/s²
So, the magnitude of the acceleration of the sled while it travels up the hill is approximately 6.13 m/s².