An object of mass M = 1,338 g is pushed at a constant speed up a frictionless inclined surface which forms an angle θ = 50 degrees with the horizontal. Incline is increasing. What is the magnitude of the force that is exerted by the inclined surface on the object?

Again, what is the normal force of weight?

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 find the magnitude of the force exerted by the inclined surface on the object, we need to analyze the forces acting on the object.

First, let's consider the weight of the object. The weight (W) is equal to the mass (M) multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s².
W = M * g

Next, we need to resolve the weight into components along the incline and perpendicular to it. The component of the weight perpendicular to the incline does not come into play because the surface is frictionless. The component of the weight parallel to the incline (W_parallel) is given by:
W_parallel = W * sin(θ)

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 (F) should be equal in magnitude and opposite in direction to the parallel component of the weight.
F = -W_parallel

Substituting the value of W_parallel, we have:
F = -W * sin(θ)

Now, we can plug in the values.
M = 1,338 g = 1.338 kg
θ = 50 degrees
g = 9.8 m/s²

First, convert the mass to kg:
M = 1,338 g = 1.338 kg

Then, find the weight:
W = M * g = 1.338 kg * 9.8 m/s²

Finally, calculate the force exerted by the inclined surface:
F = -W * sin(θ)

By substituting the values and performing the calculations, you'll get the magnitude of the force exerted by the inclined surface on the object.