physicscatalyst.com logo




Force of Friction examples




Solved examples

Question 1 A block of Mass M is moving with a velocity v on straight surface.What is the shortest distance and shortest time in which the block can be stopped if μ is coefficient of friction
a.v2/2μg,v/μg
b. v2/μg,v/μg
c.v2/2Mg,v/μg
d none of the above

Solution 1
Force of friction opposes the motion
Force of friction=μN=μmg
Therefore retardation =μmg/m=μg

From v2=u2+2as
or
S=v2/2μg

from v=u+at
or t=v/μg


Question 2A horizontal force of F N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is μ. The weight of the block is
a.μF
b. F(1+μ)
c. F/μ
d none of these

Solution 2

Let W be the weight
Reaction force=F
Weight downward=W
weight Upward=frictional force=μr=μF

For no movement
weight Upward=Weight downward
W=μF

Question 3.A body is sliding down a rough inclined plane of angle of inclination θ for which coefficient of friction varies with distance y as μ(y)=Ky where K is constant.Here y is the distance moved by the body down the plane.The net force on the body is zero at A.Find the value of constant K
a. tanθ/A
b. Acotθ
c. cottanθ/A
d. Atantanθ

Solution 3


The downward force=mgsinθ
The upward force=μmgcosθ

Net force
f(y)=mgsinθ-μmgcosθ
=mg(sinθ-kycosθ)

at y=A, f(y)=0
0=sinθ-kAcosθ
or K=tanθ/A


Question 4A given object takes n times as much times to slide down a 45 rough incline as its takes to slide down a perfectly smooth 45 incline.The coefficient of kinetic friction between the objects and incline is given by.
a. 1/(1-n2)
b.1-1/n2
c. √1/(1-n2)
d. √(1-1/n2)

Solution 4

let μ be the coefficient of friction

Acceleration is smooth 45 inclined plane=gsinθ=g/(2)sup>1/2
Acceleration in friction 45 inclined plane=gsinθ-μqcosθ=[g/(2)sup>1/2(1-μ)

Now s=ut+(1/2)at2
or 2s=at2 as =0

For smooth plane
2s=g/(2)sup>1/2t2 ---(1)

for friction plane
2s=[g/(2)sup>1/2(1-μ)(nt)2 -(2)

so (1-μ)n2=1
or μ=1-1/n2


Question 5A uniform chain of length L is lying on the horizontal surface of a table.If the coefficient of friction between the chain and the table top is μ. what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on table.?
a.L/(1+μ)
b. μL/(1+μ)
c. L/(1-μ)
d. μL/(1-μ)

Solution 5

Let m be the mass per unit length

Let L be the full length and l be the length of chain hanging

So net force downwards=mlg

Net frictional force in opposite direction=μm(L-l)g

Now mlg=μm(L-l)g

or l=μ(L-l)
or l=μL/(1+μ)



Question 6The coefficient of static and kinetic friction between a body and the surface are .75 and .50 respectively.A force is applied to the body to make it just slide with a constant acceleration which is
a. g/4
b g/2
c. 3g/4
d g

Solution 6

Minimum force with which body will just move=μsmg

After the body start moving Frictional force becomes =μkmg

So ma=μsmg-μkmg
or a=g/4


Question 7 A block of mAss M is moving with a velocity v on straight surface.What is the shortest distance and shortest time in which the block can be stopped if μ is coefficient of friction
a.v2/2μg,v/μg
b. v2/μg,v/μg
c.v2/2Mg,v/μg
d none of the above

Solution 7
Force of friction opposes the motion
Force of friction=μN=μmg
Therefore retardation =μmg/m=μg

From v2=u2+2as
or
S=v2/2μg

from v=u+at
or t=v/μg



Question 8A horizontal force of F N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is μ. The weight of the block is
a.μF
b. F(1+μ)
c. F/μ
d none of these

Solution 8

Let W be the weight
Reaction force=F
Weight downward=W
weight Upward=frictional force=μr=μF

For no movement
weight Upward=Weight downward
W=μF



link to this page by copying the following text


Class 11 Maths Class 11 Physics Class 11 Chemistry




Note to our visitors :-

Thanks for visiting our website.
DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY DISCLOSURE FOR MORE INFO.