## 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 coefficent of friction

a.v

^{2}/2μg,v/μg

b. v

^{2}/μg,v/μg

c.v

^{2}/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 v

^{2}=u

^{2}+2as

or

S=v

^{2}/2μg

from v=u+at

or t=v/μg

**Question 2**A 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 coefficent 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

b>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 4**A 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 coefficent f kinectic friction between the objects and incline is gievn by.

a. 1/(1-n

^{2})

b.1-1/n

^{2}
c. √1/(1-n

^{2})

d. √(1-1/n

^{2})

**Solution 4**
let μ be the coefficient of friction

Acceleration is smooth 45 inclined plane=gsinθ=g/(2)sup>1/2

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

Now s=ut+(1/2)at

^{2}
or 2s=at

^{2} as =0

For smooth plane

2s=g/(2)sup>1/2t

^{2} ---(1)

for friction plane

2s=[g/(2)sup>1/2(1-μ)(nt)

^{2} -(2)

so (1-μ)n

^{2}=1

or μ=1-1/n

^{2}
**Question 5**A uniform chain of length L is lying on the horizontal surface of a table.If the coefficent 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 6**The coefficent of static and kinectic 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=μ

_{s}mg

After the body start moving Frictional force becomes =μ

_{k}mg

So ma=μ

_{s}mg-μ

_{k}mg

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 coefficent of friction

a.v

^{2}/2μg,v/μg

b. v

^{2}/μg,v/μg

c.v

^{2}/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 v

^{2}=u

^{2}+2as

or

S=v

^{2}/2μg

from v=u+at

or t=v/μg

**Question 8**A 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

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