Question

Block M = 7.50 kg is initially moving up the incline and is increasing speed with
a = 3.82 m/s2 . The applied force F is horizontal. The coecients of friction
between the block and incline are s = 0.443 and k = 0.312. The angle of the incline is
25.0 degrees.
(a) What is the force F (N)?
(b) What is the normal force N (in N) between the
block and incline?
(c) What is the force of friction (N) on the block?

Answer 1

Fb = mg = 7.50kg * 9.8N/kg = 73.5N

Fb = 73.5N @ 25 deg. = Weight of box.

Fp=73.5s8in25 = 31.06N.=Force parallel to the plane(ramp).

Fv = 73.5cos25 = 66.6N = Force perpendicular to the plane(ramp).

Ff = u*Fv = 0.312 * 66.6 = 20.78N.

a. Fn = F/cos25 - Fp - Ff = ma.
1.1F - 31.06 - 20.78 = 7.5*3.82.
1.1F - 51.84 = 28.65,
1.1F = 28.65 + 51.84 = 80.49.
F = 73.2N.


b. Fv = Normal = 73.5cos25 = 66.6N.

c. Ff = u*Fv = 0.312 * 66.6 = 20.84N.