How Far Does a Block Travel Up an Incline Before Stopping?

[pulau_tiga]

Inclided plane

Hi!

I have a physics question.

A block with mass m=14.7 kg slides down an inclined plane of slope angle 44.4* with a constant velocity. It is then projected up the same plane with an initial speed 2.55 m/s. How far up the incline will the block move before coming to rest?

I know that Fnet in the x-direct (assuming x-axis is parallel to the inclined). is 0 since there is no acceleration, but constant V. Therefore Fparallel + -friction = 0. Uk = tan(44.4).

On the way back up:
Fnetx = ma
Fapp + -f + -Fgx = ma
This does not work out, as I get a to = 0.

I need acceleration in order to find the distance traveled up the incline. Using
Vf = Vi + 2ad

If anyone could help me out it would be greatly appreciated.
Thanks.

 

[Doc Al]

As you realize, on the way down the net force is zero. Use this fact to figure out the force of friction. Hint: On the way down, friction and gravity (the component parallel to the plane) act in opposite directions; on the way up, they act in the same direction. Use this fact to find the force and then the acceleration on the way up.

Also: The kinematic equation you will need is v_f^2 = v_i^2 + 2ad.

 

[wais]

Hi there!

First, let's start by finding the acceleration of the block as it slides down the inclined plane. We can use the equation Fnet = ma, where Fnet is the net force acting on the block, m is its mass, and a is the acceleration. Since the block is moving at a constant velocity, we know that the net force must be zero. This means that the force of gravity pulling the block down the incline (Fgx) must be balanced by the force of friction (Ff), which is equal to the coefficient of kinetic friction (uk) times the normal force (Fn) of the plane on the block. So we have Fgx = ukFn. We also know that Fgx = mg and Fn = mgcos(44.4), so we can substitute these values into our equation to get mg = ukmgcos(44.4). This simplifies to uk = cos(44.4), which is approximately 0.72. So the coefficient of kinetic friction is 0.72. Now we can find the acceleration by rearranging our equation to a = Fnet/m and plugging in our values for Fnet and m. This gives us a = (ukFn)/m = (0.72 x mgcos(44.4))/m = gcos(44.4) = 9.8 x cos(44.4) = 7.06 m/s^2.

Now, let's use this acceleration to find the distance traveled up the incline. We can use the equation Vf^2 = Vi^2 + 2ad, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the distance traveled. We know that Vf = 0 since the block comes to rest, Vi = 2.55 m/s, and a = 7.06 m/s^2. Plugging these values in, we get 0 = (2.55)^2 + 2(7.06)d, which simplifies to d = -1.44 m. This negative distance just means that the block will travel 1.44 m back down the incline before coming to rest.

I hope this helps! Let me know if you have any other questions. Good luck with your physics studies!