hello ok i realized what you meant. =)

An object of mass M = 921 g is pushed at a constant speed up a frictionless inclined surface which forms an angle θ = 50o with the horizontal. The normal force acts perpendicular to the slope. The slope is the x-component and y component is the normal force. (i.e. slope is the x-axis)

What is the magnitude of the force that is exerted by the inclined surface on the object?

Thanks!

To find the magnitude of the force exerted by the inclined surface on the object, we can use the concept of resolving forces into their components.

In this case, the object is being pushed up the inclined surface at a constant speed. Since there is no vertical acceleration, the normal force (N) must be equal and opposite to the gravitational force acting on the object (mg), where m is the mass of the object and g is the acceleration due to gravity.

Given that the mass of the object is 921 grams (0.921 kg) and the angle of the inclined surface is 50 degrees, we can calculate the gravitational force (mg) acting on the object:

mg = (0.921 kg) * (9.8 m/s²)
= 9.0178 N

Therefore, the magnitude of the normal force (N) is also 9.0178 N.

Since the object is being pushed up the inclined surface at a constant speed, the force exerted by the inclined surface on the object must be equal in magnitude and opposite in direction to the force exerted by the object on the inclined surface. Therefore, the magnitude of the force exerted by the inclined surface on the object is also 9.0178 N.

So, the answer is 9.0178 N.

Note: In this scenario, the force pushing the object up the inclined plane is provided by an external source, which is not explicitly mentioned in the question.