Q.63. Laws of Motion, Friction

Question 63.
A 3.70 kg block starts from rest at the top of a 30.0° incline and slides 2.10 m down the incline in 1.50 s.
1)Find the coefficient of kinetic friction between the block and the incline.
2)Find the frictional force acting on the block.
3)Find the speed of the block after it has slid a distance of 2.10 m.

Answer 63.
s = ut + 0.5at^2
=> 2.1 = 0 + 0.5a(1.5)^2
=> a = 4.2 * (4/9) = 5.6/3 m/s^2
=> net force down the plane
= mass x acceleration
= (3.7) * (5.6/3) N
= 6.91

Now, net force = downward component of weight - frictional force
=> Frictional force
= mgsin30° - 6.91
= (3.7)*(9.8)*(0.5) - 6.91
= 18.13 - 6.91
= 11.22 N.

Normal force x coefficient of friction = frictional force
=> mgcos30° * μ = 11.22
=> μ = (11.22) /[3.7*9.8*cos30°]
= 0.36

Answers:
1) 0.36
2) 11.22 N
3) v^2 = u^2 + 2as
=> v^2 = 2*(5.6/3)*(2.1) = 1.4 * 5.6
=> v = 2.8 m/s.

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