Physics question?

Question

A 5.0 kg block is sent sliding up a plane inclined at theta = thirty-seven degrees while a horizontal force F of magnitude 50 N acts on it. The coefficient of kinetic friction between the block and the plane is 0.30. How would i find the magnitude and direction of the block's acceleration? and, if i have the block's initial speed, how can i find out how far up the plane the block goes?

Answer 1

[Edit: After finishing this, I thought of one other possibility -- that the block was given an initial velocity vo going up the plane, and while moving, the 50 N horizontal force continued to be applied.

[If that's the case, you can still do the problem using the method here; the only difference is that you do not remove the horizontal force as I did below. All the same formulas apply.]

First you have to draw the inclined plane and the 37 degree angle. Somewhere up the plane, put a dot representing the 5 kg block.

Now here are the forces you have to draw in there. First, the weight of the block. That's 5.0 x 9.8 = 49 N vertically downward, and it resolves into two components -- 49 cos 37 normal (perpendicular) to the plane, and 49 sin 37 parallel to and directed the plane. (You have to draw the angles to see this.)

Next we resolve the 50 N horizontal force. Its normal component (perpendicular to the plane) is 50 cos 53, and the parallel component directed the plane is 50 sin 53. (The 37 and 53 degree angles are complementary; sin 53 = cos 37, etc.)

Now add the normal and parallel components. The total normal component is 49 cos 37 + 50 cos 53, and the parallel component directed the plane is 50 sin 53 - 49 sin 37 = 10.44 N. (Be careful of the signs here. Also note that the upward force is greater than the downward force, so the block is being pushed upward.)

The frictional force is proportional to the total force, and it acts opposite to the direction of motion. This frictional force is 0.30(49 cos 37 + 50 cos 53) = 14.7 cos 37 + 15 cos 53 = 20.77.

Here we stop. The net upward force on the block is 10.44 N, and the maximum downward frictional force is 20.77. That means the upward force is not enough to overcome friction.

The block will not move. That's your first answer.


Now let's turn to the second part of your question. This time, suppose the block was given a good, hard shove to get it going uphill at a velocity vo. Your question is, how far up the plane will it go?

The 50 N force has been removed; the block has a normal component of 49 cos 37, and a parallel component of 49 sin 37, directed back down the plane.

The frictional force is 0.30(49 cos 37) = 14.7 cos 37, directed down the plane because the block is moving upward.

The total force directed down the plane is 49 sin 37 + 14.7 cos 37 = 41.23 N.

Since F = ma, then a = -41.23 N / 5 kg = -8.25 m/s/s.

Now we turn to the formula v = at + vo, and ask how long it will take for the sliding block to come to a halt?

0 = at + vo ==> t = -vo/a and t^2 = vo^2 / a^2

To see how far up the block will slide, use

s = 1/2 at^2 + vo t = vo^2 / (2a) - vo^2 / a

s = (vo^2 / a)(1/2 - 1) = -vo^2 / (2a)

where a = -8.25 m/s/s.

There's your answer.

Answer 2

Draw a diagram of the plane, the block and all the forces acting on it. Resolve these forces into forces along and perpendicular to the plane and use this to figure the friction force. Subtract it from the compnent of the 50 N force (assuming that one is pushing the block uphill; you didn't say...!) Sum all the forces along the plane, divide by the mass of the block and you will have the acceleration a.

The distance it goes will be given by (Vo^2)/2a