2 physics questions?

Question

A 3 kg block starts from rest at the top of a 30 degree incline and slides a distance of 2m down the incline in 1.5s. Find a) the acceleration of the block, b) the coefficient of kinetic friction, c) the frictional force acting on the block, and d) the speed of the block after it has slid 2m.


Three blocks are in contact with each other on a frictionless horizontal surface. A horizontal force of 18N is applied to the first block. Find a) the acceleration of the blocks, b) the resultant force on each block, and c) the magnitude of the contact forces between the blocks. (The blocks' masses are, in order, 2kg, 3kg, 4kg)

Oh, and this isn't homework or a quiz or anything, just some concepts I'm struggling with.

Answer 1

The force pulling the block down is m*g*sinø. The force against the incline is m*g*cosø. The frictional force is then k*m*g*cosø. The net force accelerating the block is m*g*sinø - k*m*g*cosø. The acceleration is F/m, so

a = g*sinø - k*g*cosø ............ (1)

The block moved a distance s in t sec, if the acceleration was a, then

s = .5*a*t^2..............(2)

You can solve for a from (2) then use (1) to find k.

The speed of the block is a*t.



The acceleration of the blocks is a = F/m, where m = m1 + m2 + m3; To find the force on each block, assume that it its being accelerated by the blocks behind it. The end block, with mass m3, gets a force from the blocks behind equal to m3*a. The middle block gets a force that accelerates both m3 and m2, so the force it gets is a*(m2 + m3). The first block gets all the force applied.

Answer 2

To get anywhere with this problem the first thing we need to find is the speed after it has slid for 2 m.

But we need to figure out a few things first......


To find the acceleration...

a = (v2 - v1) / (t2 - t1)
a = v2 / 1.5

Since we don't know the speed of the block after 1.5 s, we can't determine the acceleration.



The second question deals with your ability to break forces down into their components.

What are the forces acting on the block?
friction
gravity
normal force

Now, we get to choose a set of axis for our system. The most logical one is the set where we use 30º as least as possible. Since we're dealing with an incline, let's use that as our coordinate system. Let the x-axis run right along the edge of the incline, and the y-axis run perpendicular to it.

This means that the normal force will go straight along the y-axis, and the frictional force will go straight along the x-axis. The only thing that splits up is gravity.

Usually, I compute the y-components before the x-components.


Forces in the y-direction...

N = Normal force (acting up)
-mg cos 30º = weight (acting down)

Since the block isn't accelerating in the y-direction, we have

N - mg cos 30º = 0
N = mg cos 30º = 25.46 N


In the x-direction...

fr = -µN (frictional force opposite the direction of motion)
mg sin 30º = (weight in the direction of motion)

The block IS accelerating, so we can set the forces equal to zero.

mg sin 30º - µN = ma

14.7 - 25.46 µ = 3a
µ = 0.5774 - 0.1178a



If we knew µ, we could calculate the frictional force. It's just

fr = µ N = 25.46 µ


We are going to have to use energy to determine the velocity. Since we're dealing with force, energy is being removed from the system. We have the following equation to guide us.

Wnc = Ef - Ei

The work done by non-conservative forces is equal to the change in energy. Initially the block starts with KE = 0 and ends up with PE = 0.

The inital height of the object is h = 2 * sin 30º = 1m

The work done by friction, a non-conservative force, is equal to -µNx, where x is the distance the object traveled (2 in this case).

Wnc = KEf - PEi
-50.92µ = 1.5 v² - 29.4
v² = 19.6 - 33.947µ

But µ = 0.5774 - 0.1178a, so

v² = 19.6 - 33.947(0.5774 - 0.1178a)

and a = v / 1.5

v² = 19.6 - 33.947(0.5774 - 0.0785v)
Solving for v = 2.664 m/s.

Now we know v, we can find

a = 1.776 m/s²
µ = 0.3682
fr = 9.374 N
v = 2.664 m/s


This was an extremely difficult problem, and probably not well suited for testing out some difficult concepts. I deliberately skipped around a bit to make the answer shorter, but I hope it is still clear to you what I did. I fear that my answer will be too confusing due to the nature of the problem. If we had been given the acceleration or the speed to begin with, it would have been a much simpler problem.


I don't have time to finish your second problem. I may come back to it later. Good luck!!




*EDIT* I see what the person before me did.

s = ½at²

He is correct, and will make the problem much more simpler than how I worked it out. I'll leave my original posted because I feel I explained the forces situation pretty well. Kudos to him!