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?
What is the normal component of weight?
ok, so what do you mean by component of weight? anyways, this is what i thought should be the answer F= mgsin(θ) is that correct?
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 break down the forces acting on the object.
First, let's consider the weight of the object. The weight is given by the equation:
Weight = mass * gravitational acceleration
Given that the mass of the object is 1,338 g or 1.338 kg, and the standard gravitational acceleration is 9.8 m/s², we can calculate the weight:
Weight = 1.338 kg * 9.8 m/s²
Next, let's consider the force acting along the incline. This force can be broken down into two components: the force pushing the object up the incline and the force pulling the object down the incline.
The force pushing the object up the incline is equal to the force of gravity pulling the object down the incline. This force is given by the equation:
Force_up = Weight * sin(theta)
where theta is the angle the incline forms with the horizontal (50 degrees in this case).
Finally, the magnitude of the force exerted by the inclined surface on the object is equal to the force pushing the object up the incline. So, we have:
Magnitude_of_force = Force_up
Therefore, to find the magnitude of the force, plug in the values:
Magnitude_of_force = Weight * sin(theta)
Magnitude_of_force = (1.338 kg * 9.8 m/s²) * sin(50 degrees)