M*g = 2.2 * 9.8 = 21.56 N. = Wt. of block.
Fp = 21.56*sin27.7 = 10.02 N. = Force
parallel to incline.
Fn = 21.56*Cos27.7 = 19.09 N. = Normal
force.
Fk = u*Fn = 0.41 * 19.09 = 7.83 N. = Force of kinetic friction.
a. a = (Fp-Fk)/M = (10.02-7.83)/2.2 = 0.995 m/s^2
b. L = h/sin27.7 = 3.6/sin27.7 = 7.74 m.
= Length of ramp.
V^2 = Vo^2 + 2a*L = 0 + 1.99*7.74 = 15.4
V = 3.92 m/s.
Can someone help me solve this one? I've tried it a few times, but I don't think I'm getting it right.
A block of mass M = 2.2 kg starts from rest at a height of 3.6 m on a fixed inclined plane. The coefficient of kinetic friction between the block and the incline is μ = 0.41. The incline plane makes an angle θ = 27.7 ° with the horizontal.
a) What is in m/s2 the acceleration of the block down the ramp?
b) What is in m/s the speed of the block at the bottom of the ramp?
The block continues to slide on the ground with the same coefficient of friction.
c) What is in m/s2 the acceleration of the block on the ground?
d) how far will the block slide on the ground until coming to rest?
1 answer
9 answers