A 1560 rocket is to be launched with an initial upward speed of 53.0 . In order to assist its engines, the engineers will start it from rest on a ramp that rises 53 above the horizontal (the figure ). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of 2000 , and friction with the ramp surface is a constant 500 .

MPH, FPS, M/S, lbs, kg, ????

23.4 To find the net force acting on the rocket, we need to consider the forces involved. In this case, we have three forces acting on the rocket:

1. Weight (gravitational force): This force acts vertically downwards and is given by the equation F_weight = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. Normal force: This force acts perpendicular to the ramp and counteracts the weight of the rocket. Since the rocket is on an inclined ramp, the normal force can be calculated using the equation F_normal = m * g * cos(theta), where theta is the angle of inclination (53 degrees in this case).

3. Frictional force: This force opposes the motion of the rocket along the ramp. The magnitude of the frictional force can be given by the equation F_friction = mu * F_normal, where mu is the coefficient of friction between the ramp and the rocket, and F_normal is the normal force.

Now, let's calculate the forces:

1. Weight:
F_weight = m * g
= 1560 kg * 9.8 m/s^2
≈ 15288 N

2. Normal force:
F_normal = m * g * cos(theta)
= 1560 kg * 9.8 m/s^2 * cos(53 degrees)
≈ 9798 N

3. Frictional force:
F_friction = mu * F_normal
= 500 * 9798 N
= 4899000 N

Since the rocket is launched vertically, we only need to consider the forces acting in the vertical direction. The net force acting on the rocket in the vertical direction is given by:

Net force = Sum of vertical forces
= F_weight - F_normal
= 15288 N - 9798 N
≈ 5489 N (upward)

Therefore, the net force acting on the rocket in the vertical direction is approximately 5489 N, directed upward.