Alistar places a 6.4 kg block on an incline plane that is 23 degrees above the horizontal. The block starts from rest, and in 2.2 seconds, travels 1.4 meters. What is the force of friction on the block?

fh= 6.4×23

fh=147.2

fv=2.2×1.4

fv=3.08

f=147.2+3.08

f=150.28

1.28m/s

Explanation:

We are given: m=6.4 kg,θ=23°,x=1.4 m,t=2.2 s

(a) At constant acceleration

   xf=vit+21at2

solving

   a=t22(xf−vit)=(2.2s)22(1.4m−0)=0.58m/s2

From the acceleration, we can calculate the friction force, answer (c), next.

We are given: m=3.00kg,θ=30.00,x=2.00m,t=1.50s

(a) At constant acceleration

   xf=vit+21at2

solving

   a=t22(xf−vit)=(1.50s)22(2.00m−0)=1.78m/s2

From the acceleration, we can calculate the friction force, answer (c), next.

(c) Take the positive x axis down parallel to the incline, in the direction of the acceleration. We apply Newton’s second law:

   ∑Fx=mgsinθ−f=ma

Solving,    f=m(gsinθ−a)

Substituting,

    f=(6.4)[(9.80)sin23°-0.58] = 20.79 N

(b) Applying Newton’s law in the y direction (perpendicular to the incline), we have no burrowing-in or taking-off motion. Then the y component of acceleration is zero:

    ∑Fy=n–mgcosθ=0

Thus n=mgcosθ

Because  f=μkn

we have  μk=mgcosθf=20.79/((6.4)(9.80)cos23°)=0.360

(d) vf=vi+at so vf=0+(1.78m/s2)(1.50s)=2.67m/s

0+(0.58)(2.2)= 1.28m/s

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