Posted by **carlton** on Monday, October 17, 2011 at 10:24am.

A 450 g block is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.670, and the coefficient of kinetic friction is 0.380. A force of magnitude P pushes the block forward and downward as shown in the figure below. Assume the force is applied at an angle of 39.0° below the horizontal.

(a) Assuming P is large enough to make the block move, find the acceleration of the block as a function of P. (Use P as necessary and round numbers to the third decimal place.) (b) If P = 6.81 N, find the acceleration and the friction force exerted on the block.

a = m/s2

Ffriction = N

(c) If P = 13.62 N, find the acceleration and the friction force exerted on the block.

a = m/s2

Ffriction = N

(d) Describe in words how the acceleration depends on P.

- physics -
**Henry**, Tuesday, October 18, 2011 at 4:33pm
Wb = mg = 0.45kg * 9.8N/kg = 4.41N. =

Weight of block.

Fb = (4.41N,0deg.).

Fp=Fh = 4.41sin(0) = 0 = Force parallel

to hor. surface.

Fv = 4.41cos(0) = 4.41N. = Force perpendicular to hor. surface.

Ff = u*Fv = 0.67 * 4.41 = 2.95N. = Force of static friction.

Ff = u*Fv = 0.360 * 4.41 = 1.59N. = Force of kinetic friction.

a. P must overcome the static friction to move the block:

Pcos39 - Ff = 0,

Pcos39 = Ff,

Pcos39 = 2.95,

P = 2.95 / cos39 = 3.8N. = Force applied.

When block starts to move, P must overcome kinetic friction to accelerate:

Fn = Pcos39 - Ff = 2.95 - 1.59 = 1.36 = Net force.

a = Fn / m = 1.36 / 0.45 = 3.03m/s^2.

b. Fn = 6.81cos39 - 1.59 = 3.7N.

a = Fn / m = 3.7 / 0.45 = 8.23m/s^2.

c. Fn = 13.62cos39 - 1.59 = 8.99N.

a = 8.99 / 0.45 = 19.98m/s^2.

d. Increasing P causes an increase in the net force, and the acceleration is proportional to the net force(a=Fn/m).

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