Posted by waqas on Saturday, August 6, 2011 at 9:29am.
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?

physics  Henry, Sunday, August 7, 2011 at 3:50pm
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.
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