# Why is acceleration constant?

• simpleton

#### simpleton

Hi,

I just saw the solution to a problem I was pondering about recently and did not really understand one of the steps/assumptions they made, and therefore I hope that you guys can clarify my doubts.

The question is like this, you have a block on a frictionless inclined slope of height h and inclined x from the horizontal. The slope is on a frictionless surface as well. The block has mass m and the slope has mass M. Calculate the time taken for the block to reach the bottom of the slope.

For those who wants to know, the answer is:

sqrt[(2h(M + m sin^2 x))/((M+m) g sin^2 x)]

But I have a few doubts about the method used to solve the problem (I am talking about the method that uses Newtonian Mechanics, not Lagrangian, my physics is not that good yet XD).

In the method given, You draw the forces exerted on the block and on the slope. For the block, the forces are mg (weight) and N (reaction force from the slope). For the slope, it is N(the reaction force from the block). Then you assume that the reaction force N is constant, so after you applied Newton's Laws and resolved the accelerations into the correct axes, you can use kinematics equations.

However, I don't understand why N is constant throughout the motion. Isn't the slope moving away from the block? If it is, then N should change rit? Actually, when I was trying to solve the problem at first, I thought that there might be moments when the block might be not in contact with the slope, so I was thinking of the block making repeated elastic collisions with the slope, but it seems I am wrong.

Therefore, can someone please explain to me why N is constant, or give a formal proof? Thanks a lot.

Hi,
However, I don't understand why N is constant throughout the motion. Isn't the slope moving away from the block?
Therefore, can someone please explain to me why N is constant, or give a formal proof? Thanks a lot.

Why do you think the "slope should move away". The block is in contact with the surface all the time and since the weight of the block is constant and I assume also that the slope is constant. Then at any time and position of the downward slide you should calculate the same normal force.

I think the slope should move away because the block is exerting a force onto the slope, which pushes the slope in the opposite direction. Therefore, the slope is moving away from the block and thus the reaction force should change.

I think the slope should move away because the block is exerting a force onto the slope, which pushes the slope in the opposite direction. Therefore, the slope is moving away from the block and thus the reaction force should change.

When you stand on your feet does the Earth move away from your feet?

Sorry i misunderstood your problem since i am tired (doing homework at 3am). But what i said still holds. They assume that the normal is constant since the weight is constant and the slope does not change. If for example the slide would be a parabola the acceleration of the incline would not be constant.

Sorry, I don't understand why the fact that the slope and the weight does not change will make the normal force constant. Because after the block has moved, the velocities of the block and the slope will be different. And from what I understand, the reaction force is the force at which the electrons repel you when you go into the microscopic level rit? So if you push harder onto a wall or something, there is a greater normal force because the repulsion will be greater.

In this case, the slope now has velocity and is moving away from the block. If it is moving away from the block, shouldn't the normal force (repulsion of the electrons) decrease too, because as the slope is moving away from you, the force at which you push against the slope will decrease?

In this case, the slope now has velocity and is moving away from the block. If it is moving away from the block, shouldn't the normal force (repulsion of the electrons) decrease too, because as the slope is moving away from you, the force at which you push against the slope will decrease?

The problem states that everything is frictionless so there is not much use for thinking of electromagnetic forces.

The thing is that the block is also moving towards the slope (although a different region of the slope) it falls because of its weight so at any time the distance between the block and the surface of the slope is the same (for the purpose of this problem the distance is 0). So the normal force can be considered constant.

Hmm, I guess that does make sense.

However, why then is the normal force not mg*cos x? If the slope is stationary can cannot move, that would have been the normal reaction force. Why is the normal force not mg*cos x if the slope can move (I think it will be smaller than mg*cos x)?

However, why then is the normal force not mg*cos x?
Because the "weight" (downwards force) of the block = m(g-a), where a is the vertical rate of acceleration of the block (note both g and a should be negative values).

Consider yourself on that moving inclined plane(non inertial frame)
now you will observe a normal force on the block which remains constant
Therefore during the motion the normal force on the block remains constant(inertial frame)