Question #95734

A 3.95-kg block starts from rest at the top of a 30.0° incline and slides a distance of 1.80 m down the incline in 1.90 s.
(a) Find the magnitude of the acceleration of the block.

(b) Find the coefficient of kinetic friction between block and plane.

(c) Find the friction force acting on the block.

(d) Find the speed of the block after it has slid 1.80 m.

Expert's answer

s=12at2=1.805×a=1.80a=1 m sec2s=\frac{1}{2}at^2=1.805\times a=1.80\\a=1\ m\ sec^{-2}

(a) = magnitude of accelration = 1 m sec21 \ m\ sec^{-2}


(b)

a=gsin30°μgcos30°a=g\sin30\degree -\mu g\cos30\degree =1=1

μ=gsin30°1gcos30°=453=0.46188\mu=\frac{g\sin30\degree-1}{g\cos30\degree}=\frac{4}{5\sqrt3}=0.46188


(c)

Ff=μmgcos30°=453×3.95×10×32=15.8 NF_f=\mu mg\cos30\degree=\frac{4}{5\sqrt3}\times3.95\times10\times \frac{\sqrt3}{2}=15.8\ N


(d)

speed of block is given by:

v=2as=2×1×1.8=1.897 m sec1v=\sqrt{2as}=\sqrt{2\times 1\times 1.8}=1.897\ m\ sec^{-1}


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