An object of mass M = 924 g is pushed at a constant speed up a frictionless inclined surface which forms an angle è = 50 degrees 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?

There is an F shown, at an angle in the figure.

That angle is important.

The direction of the applied force is important, and so it the angle of the incline. Apply vertical and horizontal forces balance equations to solve.

Thank you

To determine the magnitude of the force exerted by the inclined surface on the object, we can use Newton's second law of motion. This law states that the net force acting on an object is equal to the product of its mass and acceleration.

First, we need to find the acceleration of the object. Since the object is moving at a constant speed, the net force acting on it must be zero. However, there are two forces acting on the object on the inclined surface: its weight (mg) acting straight downward and the force exerted by the inclined surface pushing the object upward.

Using trigonometry, we can find the component of the weight that is parallel to the inclined surface. The weight's component parallel to the inclined surface is given by mg * sin(θ), where θ is the angle of the incline (θ = 50 degrees). This is the force exerted by gravity that tends to pull the object down the slope.

Since the object is moving at a constant speed, the force exerted by the inclined surface must counterbalance the component of the weight parallel to the incline. Therefore, the magnitude of the force exerted by the inclined surface on the object is also mg * sin(θ).

Now, we can calculate the magnitude of the force:
mass (m) = 924 g = 0.924 kg
gravitational acceleration (g) = 9.8 m/s^2 (standard value)
angle of incline (θ) = 50 degrees

Force exerted by the inclined surface = 0.924 kg * 9.8 m/s^2 * sin(50 degrees)
= 0.924 kg * 9.8 m/s^2 * 0.766
≈ 7.093 N

Therefore, the magnitude of the force exerted by the inclined surface on the object is approximately 7.093 N.