An object of mass M = 1,284 g is pushed at a constant speed up a frictionless inclined surface which forms an angle θ = 50 degrees with the horizontal. What is the magnitude of the force that is exerted by the inclined surface on the object?

(I'm using 1.284*9.81*cos50 to get the force but I'm still getting a wrong answer.)

Can anyone explain the steps into solving this kind of problem.

The normal force depends upon the angle the pushing force is applied to the block. If it is applied parallel to the inclined surface (along the direction of motion), I agree with your equation. Your answer would be in Newtons.

Other force application directions are possible.

Tilt your axis such that, Y is pointing to the opposite of Weight of M. And X is pointing towards the right. (like usual). [The axis shown in the diagram requires you to find Fnet, which I don't know how to find; thus use this method].

In that case,

Positive being up,

-W(of M)+Ncos50=0
=> N=Mg/cos50

plug your numbers, that's your answer.

To solve this problem, you need to apply the principles of Newton's second law and decompose forces into components parallel and perpendicular to the inclined surface.

1. Start by drawing a diagram of the situation. Label the object's weight (mg) pointing straight down, and draw the inclined surface's normal force (N) perpendicular to the surface, and the force vector exerted by the inclined surface (F).

2. Recognize that the weight vector can be divided into two components: one parallel to the inclined surface and one perpendicular to it. The component parallel to the inclined surface is given by mg*sinθ (where θ is the angle of the incline), while the component perpendicular to the inclined surface is mg*cosθ.

3. In this problem, the object is pushed at a constant speed, implying that there is no net force in the direction of motion. The force exerted by the inclined surface (F) must balance the component parallel to the inclined surface (mg*sinθ).

4. The magnitude of the force exerted by the inclined surface can be found by setting up the equation F = mg*sinθ.

Applying this approach to the given values:
- Mass (M) = 1,284 g = 1.284 kg
- Angle (θ) = 50 degrees

First, convert the mass to kilograms: M = 1.284 kg.

Then, calculate the force parallel to the inclined surface: F_parallel = M * g * sinθ = 1.284 kg * 9.81 m/s^2 * sin(50 degrees) = 6.30 N.

So, the magnitude of the force exerted by the inclined surface on the object is 6.30 N.

It is essential to use the sine function for the parallel component because the force acts parallel to the inclined surface. Double-check your calculations to ensure accuracy.