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A 3.08 kg block starts from rest at the top of a 30° incline and accelerates uniformly down the incline, moving 2.28 m in 1.60 s.

Find the magnitude of the acceleration of the block.

Find the coefficient of kinetic friction between the block and the incline.

Find the magnitude of the frictional force acting on the block.

Find the speed of the block after it has slid a distance 2.28 m.

2007-10-15 10:59:09 · 1 answers · asked by John L 1 in Science & Mathematics Physics

1 answers

The equations for motion under constant acceleration are:

D = Vi x T + (1/2) x A x T^2
Vf = Vi + A x T

The first gives the distance travelled in terms of initial velocity Vi, acceleration A, and time T. The second gives the final velocity.

You first want to find A, and you know Vi (rest means 0), distance, and time so you can use the distance equation.

With A, you can compute the final velocity for the last part of the question.

Next comes friction and resolution of forces. There are three forces acting on the block:

the Normal force exerted by the incline on the block, perpendicular to the incline.

the Frictional force along the incline

and the gravitational force or W.

W is given by mass times acceleration, where M is the mass of the block and the acceleration is the acceleration due to gravity, or 9.8 m/s^2.

Knowing the angle of the incline, we can split the weight into two components: Wn is the component normal to the incline, and Wp is the component parallel to the incline.

Since the block stays on the incline, the sum of the forces in the normal direction is 0:

0 = Wn + N

Since the block accelerates down the incline, the sum of the forces parallel to the incline are mass x acceleration:

M x A = Wp + F

We know M, A, W and the angle so we can compute Wp, Wn, F, and N.

Now we have the equation connecting N and F.

F = N x Ck

where Ck is the coefficient of kinetic friction. Since we have F and N, we can compute Ck to get the answer to part 2.

(Note that some of the forces will come out negative. This is to be expected. Forces parallel to the plane and down should be positive; forces up the plane should be negative, etc.)

2007-10-17 17:54:22 · answer #1 · answered by simplicitus 7 · 0 0

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