1. A 10.0 kg block is on an incline plane of variable angle. The coefficients of

Question

1. A 10.0 kg block is on an incline plane of variable angle. The coefficients of static and kinetic friction are ?s=0.29 and ?k=0.22.
(a) What is the minimum angle such that the block slides down the plane?
_____ o

(b) What is the frictional force at this angle?
_____ N

(c) If the angle of the incline plane is half of that in part (a) what are the force of friction and acceleration of the block down the plane if it released from rest?
______ N

_______ m/s2

(d) If the angle of the incline plane is twice that in part (a) what are the force of friction and acceleration of the block down the plane if it released from rest?
_____N

______m/s2

(e) Suppose the block is given an initial velocity up the incline plane when the angle of the plane is half of that in part a. What is the force of friction and acceleration of the block as it moves up the plane? What is the force of friction and acceleration of the block after it reaches its maximum height on the plane?
_____N

______m/s2

_________ N

_____m/s2

(f) Suppose the block is given an initial velocity up the incline plane when the angle of the plane is twice that in part a. What is the force of friction and acceleration of the block as it moves up the plane? What is the force of friction and acceleration of the block after it reaches its maximum height on the plane?

_______N

_____m/s2

_______ N

_______ m/s2

2. You are sitting in the back of a produce truck that has stopped at a red light. You want to measure its acceleration when the light turns green. You could actually measure the acceleration of the truck by measuring the increase in weight of a mass with a spring scale. You decide to hang a large potato on a spring scale. It weighs 4.6N. The light turns green. The truck begins to accelerate. As the truck accelerates apparent weight of the large potato increases to 5.03N as it swings from the vertical.
(a) To what angle does the potato swing from the vertical?
_______ o
(b) What is the acceleration of the truck?
_________ m/s/s

Explanation / Answer

1) a)by angle of repose= for the block to slide down -tan theeta>static friction

also newton's second law tells you that

sum of forces = ma

the forces acting on the skier are the component of gravit down the plane and the force of friction acting up the plane

the component of grav down the plane is mgsin(theta)
the force of friction up the plane is u mg cos(theta) where u is the coefficient of friction

so we have

mg sin(theta) - u mg cos(theta) = ma

motion will be initiated when the left side exceeds zero, or when

g(sin(theta)- u cos(theta)>0 or when
sin(theta) > u cos (theta)

div through by cos (theta) and find that motion will begin when

sin(theta/cos(theta)> u

but the left is just tan(theta), so our condition is

tan(theta)>mu (coeff of friction)

so the block will slide down a slope where
therfore theeta(angle)=tan(inverse of 0.29) or angle>16.2 degree

b)the frictional force acting will be coeef of friction (u)* normal

normal=umgcos theta

0.22*10*10(g)*cos theta=22cos 16.2N

c)at this point frictional force will be equal to mgsin theta

10*10*sin 8.1=100sin 8.1 N

acceleration will be 0 as system will be in equilibrium

d)force of friction will be same as kinetic friction which is umgcos theta which will be 22cos 32.4N

acc. will be (mgsin theta -f)/m=a

(100sin 32.4 - 22cos32.4)/10

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