An object of mass M = 911 g is pushed at a constant speed up a frictionless inclined surface which forms an angle θ = 50o with the horizontal as shown in the figure below. What is the magnitude of the force that is exerted by the inclined surface on the object?

Since there is zero acceleration, the net force on the object is zero.

You do not say what the angle of the applied force (with respect to the incline) is. Perhaps the "figure below" (which you do not provide) shows it. The answer depends upon it.

To find the magnitude of the force exerted by the inclined surface on the object, we need to consider the forces acting on the object. In this case, there are two forces acting on the object: the force of gravity and the normal force exerted by the inclined surface.

1. Start by considering the force of gravity acting on the object. The force of gravity can be calculated using the formula:

Force of gravity = mass × acceleration due to gravity

The acceleration due to gravity is approximately 9.8 m/s^2.

Force of gravity = 0.911 kg × 9.8 m/s^2

2. Decompose the force of gravity into two components: one parallel to the incline and one perpendicular to the incline. The parallel component is given by:

Force parallel = Force of gravity × sin(θ)

where θ is the angle between the incline and the horizontal.

Force parallel = (0.911 kg × 9.8 m/s^2) × sin(50°)

3. Since the object is moving at a constant speed, the net force acting on it must be zero. Therefore, the force exerted by the inclined surface (the normal force) must balance the force of gravity. The magnitude of the force exerted by the inclined surface is equal to the perpendicular component of the force of gravity:

Force perpendicular = Force of gravity × cos(θ)

Force perpendicular = (0.911 kg × 9.8 m/s^2) × cos(50°)

So, the magnitude of the force exerted by the inclined surface on the object is equal to the force perpendicular:

Force exerted by the inclined surface = (0.911 kg × 9.8 m/s^2) × cos(50°)