Laws of Motion- 3 Practice Questions - DPP for NEET

 Page 1


(1) (a)  Let the contact force on the block by the surface be F
which makes an angle q with the vertical. The component
of F perpendicular to the contact surface is the normal
force N and the component F parallel to the surface is the
friction f. As the surface is horizontal, N is vertically upward.
For vertical equilibrium
F
N
f
q
N = Mg = (0.400) (10) = 4.0 N
The frictional force is f = 3.0 N
tan q = 
f
N
 = 
3
4
 Þ q = tan
–1
 (3/4)  = 37º
(2) (c) The magnitude of the contact force is
    F = 
N f
2 2
+
 = 
( ) () 4 3
2 2
+
 = 5.0 N
(3) (c) The forces on the block are
N
mg
f
q
(i) the weight mg downward by the earth
(ii) the normal contact force N by the incline, and
(iii) the friction f parallel to the incline up the plane, by the
incline.
As the block is at rest, these forces should add up to zero.
Also since q is the maximum angle to prevent slipping, this
is a case of limiting equilibrium and so
f = m
S
N
Taking component perpendicular to the lncline,
N – mg cos q = 0 Þ N = mg cos q ....... (1)
Taking component parallel to the incline
f – mg sin q = 0 Þ f = mg sin q   ........ (2)
\  m
S
N = mg sin q
Dividing (2) by (1) m
s
 = tanq
      q = tan
–1
m
S
 = tan
–1
 (0.3)
(4) (a) When the maximum force F is applied, both the blocks
move together towards right. The only horizontal force on
the upper block of mass m is that due to the friction by the
lower block of mass M. Hence this force on m should be
towards  right. The force of friction on M by m should be
towards left by Newton's third law . As we are talking of the
minimum possible force F that can be applied, the friction
is limiting and hence
f = mN, where N is normal force.
in the vertical direction, there is no acceleration
\ N = mg
in the horizontal direction,
let the acceleration be a, then
mN = ma
mmg = ma
a = mg          
f = N m
Next consider the motion of M
The equation of motion is
F = mN = Ma
F – mmg = Mmg
F = mg (M + m)
f = N m F
N
Mg
(5) (b) When A moves with B the force opposing the motion is
the only force of friction between B and S the horizontal
and velocity of the system is constant
B
R
1
A
R
2
F
f
1
F = f
1
 = mR
1
 = 0.25 (4 + 8) = 3N
(6) (d) When A is held stationary the friction opposing the
motion is between A and B and B and S. So
B
R
1
R
2
F
f
1
A
S
f
2
F = mR
1
 + mR
2
 = 3 + 0.25 (4)
F = 3 + 1 = 4 N
Page 2


(1) (a)  Let the contact force on the block by the surface be F
which makes an angle q with the vertical. The component
of F perpendicular to the contact surface is the normal
force N and the component F parallel to the surface is the
friction f. As the surface is horizontal, N is vertically upward.
For vertical equilibrium
F
N
f
q
N = Mg = (0.400) (10) = 4.0 N
The frictional force is f = 3.0 N
tan q = 
f
N
 = 
3
4
 Þ q = tan
–1
 (3/4)  = 37º
(2) (c) The magnitude of the contact force is
    F = 
N f
2 2
+
 = 
( ) () 4 3
2 2
+
 = 5.0 N
(3) (c) The forces on the block are
N
mg
f
q
(i) the weight mg downward by the earth
(ii) the normal contact force N by the incline, and
(iii) the friction f parallel to the incline up the plane, by the
incline.
As the block is at rest, these forces should add up to zero.
Also since q is the maximum angle to prevent slipping, this
is a case of limiting equilibrium and so
f = m
S
N
Taking component perpendicular to the lncline,
N – mg cos q = 0 Þ N = mg cos q ....... (1)
Taking component parallel to the incline
f – mg sin q = 0 Þ f = mg sin q   ........ (2)
\  m
S
N = mg sin q
Dividing (2) by (1) m
s
 = tanq
      q = tan
–1
m
S
 = tan
–1
 (0.3)
(4) (a) When the maximum force F is applied, both the blocks
move together towards right. The only horizontal force on
the upper block of mass m is that due to the friction by the
lower block of mass M. Hence this force on m should be
towards  right. The force of friction on M by m should be
towards left by Newton's third law . As we are talking of the
minimum possible force F that can be applied, the friction
is limiting and hence
f = mN, where N is normal force.
in the vertical direction, there is no acceleration
\ N = mg
in the horizontal direction,
let the acceleration be a, then
mN = ma
mmg = ma
a = mg          
f = N m
Next consider the motion of M
The equation of motion is
F = mN = Ma
F – mmg = Mmg
F = mg (M + m)
f = N m F
N
Mg
(5) (b) When A moves with B the force opposing the motion is
the only force of friction between B and S the horizontal
and velocity of the system is constant
B
R
1
A
R
2
F
f
1
F = f
1
 = mR
1
 = 0.25 (4 + 8) = 3N
(6) (d) When A is held stationary the friction opposing the
motion is between A and B and B and S. So
B
R
1
R
2
F
f
1
A
S
f
2
F = mR
1
 + mR
2
 = 3 + 0.25 (4)
F = 3 + 1 = 4 N
32
DPP/ P 11
(7) (d) In this situation for dynamic equilibrium of B
R
1
R
2
F
f
1
A
f
2
f
2
F = mR
1
 + mR
2
 + T ..... (1)
While for the uniform motion of A
T = mR
2
....... (2)
Substituting T from Equation (2) in (1) we get
F = mR
1
 + 2mR
2
 = 3 + 2 x 1 = 5N
(8) (a) Figure shows the forces acting on the two blocks. As
we are looking for the maximum value of M/m, the
equilibrium is limiting. Hence the frictional forces are equal
to m times corresponding normal force.
Equilibrium of the block m gives
M
m
q
N
1
mN
1
mg
T
mN
1
mg
N
2
T = mN
1
 and N
1
 = mg Þ T = mmg .... (1)
Next consider the equilibrium of the block M. Taking
components parallel to the incline
T + mN
2
 = Mg sin q
Taking components normal to the Incline
N
2
 = Mg cos q
These give T = Mg (sin q – m cos q) ...... (2)
From (1) and (2) mmg = Mg (sin q – m cos q)
M
m sin cos
m
=
q-mq
(9) (a) The situation is shown in figure in the limiting
equilibriums the frictional force f will be equal to mN.
f = N
S
m
F
N
Mg
F
q
For horizontal equilibrium
F sin q = mN
For vertical equilibrium
F cos q + mg = N
Eliminating N from these equations
F sin q = mF cos q + mmg
F = 
) cos (sin q m - q
m
(10) (c) If y is the maximum length of chain which can be hang
out side the table without sliding, then for equilibrium of
the chain, the weight of hanging part must be balanced by
force of friction from the portion on the table
y
W
W’
L–y
f
1
R
W = f
L
....... (1)
But from figure W = 
M
L
y g and
R = W' = 
M
L
(L – y)g
So that f
L
 = mR = 
mM
L
(L – y) g
Substituting these values of W and f
L
 in equation (1) we
get
M
L
yg = m 
M
L
(L – y) g
(11) (d) The insect will crawl up the bowl till the component of
its weight along the bowl is balanced by limiting friction
so, resolving weight perpendicular to the bowl and along
the bowl we get
R
y
f
L
q
mg sinq
mg cosq
R = mg sin
q
f
L
 = mg cos q
tanq =
m
=
m
=
1
R
R
f
R
1
1
L
  ;  
m
=
-
1
y R
y
2 2
m
2
y
2
 = R
2
 – y
2   
;  
 
  y =
1
R
2
1
+ m
So, h = R – y = R –
1
R
2
+ m
= R
ú
ú
û
ù
ê
ê
ë
é
+ m
-
) 1 (
1
1
2
Page 3


(1) (a)  Let the contact force on the block by the surface be F
which makes an angle q with the vertical. The component
of F perpendicular to the contact surface is the normal
force N and the component F parallel to the surface is the
friction f. As the surface is horizontal, N is vertically upward.
For vertical equilibrium
F
N
f
q
N = Mg = (0.400) (10) = 4.0 N
The frictional force is f = 3.0 N
tan q = 
f
N
 = 
3
4
 Þ q = tan
–1
 (3/4)  = 37º
(2) (c) The magnitude of the contact force is
    F = 
N f
2 2
+
 = 
( ) () 4 3
2 2
+
 = 5.0 N
(3) (c) The forces on the block are
N
mg
f
q
(i) the weight mg downward by the earth
(ii) the normal contact force N by the incline, and
(iii) the friction f parallel to the incline up the plane, by the
incline.
As the block is at rest, these forces should add up to zero.
Also since q is the maximum angle to prevent slipping, this
is a case of limiting equilibrium and so
f = m
S
N
Taking component perpendicular to the lncline,
N – mg cos q = 0 Þ N = mg cos q ....... (1)
Taking component parallel to the incline
f – mg sin q = 0 Þ f = mg sin q   ........ (2)
\  m
S
N = mg sin q
Dividing (2) by (1) m
s
 = tanq
      q = tan
–1
m
S
 = tan
–1
 (0.3)
(4) (a) When the maximum force F is applied, both the blocks
move together towards right. The only horizontal force on
the upper block of mass m is that due to the friction by the
lower block of mass M. Hence this force on m should be
towards  right. The force of friction on M by m should be
towards left by Newton's third law . As we are talking of the
minimum possible force F that can be applied, the friction
is limiting and hence
f = mN, where N is normal force.
in the vertical direction, there is no acceleration
\ N = mg
in the horizontal direction,
let the acceleration be a, then
mN = ma
mmg = ma
a = mg          
f = N m
Next consider the motion of M
The equation of motion is
F = mN = Ma
F – mmg = Mmg
F = mg (M + m)
f = N m F
N
Mg
(5) (b) When A moves with B the force opposing the motion is
the only force of friction between B and S the horizontal
and velocity of the system is constant
B
R
1
A
R
2
F
f
1
F = f
1
 = mR
1
 = 0.25 (4 + 8) = 3N
(6) (d) When A is held stationary the friction opposing the
motion is between A and B and B and S. So
B
R
1
R
2
F
f
1
A
S
f
2
F = mR
1
 + mR
2
 = 3 + 0.25 (4)
F = 3 + 1 = 4 N
32
DPP/ P 11
(7) (d) In this situation for dynamic equilibrium of B
R
1
R
2
F
f
1
A
f
2
f
2
F = mR
1
 + mR
2
 + T ..... (1)
While for the uniform motion of A
T = mR
2
....... (2)
Substituting T from Equation (2) in (1) we get
F = mR
1
 + 2mR
2
 = 3 + 2 x 1 = 5N
(8) (a) Figure shows the forces acting on the two blocks. As
we are looking for the maximum value of M/m, the
equilibrium is limiting. Hence the frictional forces are equal
to m times corresponding normal force.
Equilibrium of the block m gives
M
m
q
N
1
mN
1
mg
T
mN
1
mg
N
2
T = mN
1
 and N
1
 = mg Þ T = mmg .... (1)
Next consider the equilibrium of the block M. Taking
components parallel to the incline
T + mN
2
 = Mg sin q
Taking components normal to the Incline
N
2
 = Mg cos q
These give T = Mg (sin q – m cos q) ...... (2)
From (1) and (2) mmg = Mg (sin q – m cos q)
M
m sin cos
m
=
q-mq
(9) (a) The situation is shown in figure in the limiting
equilibriums the frictional force f will be equal to mN.
f = N
S
m
F
N
Mg
F
q
For horizontal equilibrium
F sin q = mN
For vertical equilibrium
F cos q + mg = N
Eliminating N from these equations
F sin q = mF cos q + mmg
F = 
) cos (sin q m - q
m
(10) (c) If y is the maximum length of chain which can be hang
out side the table without sliding, then for equilibrium of
the chain, the weight of hanging part must be balanced by
force of friction from the portion on the table
y
W
W’
L–y
f
1
R
W = f
L
....... (1)
But from figure W = 
M
L
y g and
R = W' = 
M
L
(L – y)g
So that f
L
 = mR = 
mM
L
(L – y) g
Substituting these values of W and f
L
 in equation (1) we
get
M
L
yg = m 
M
L
(L – y) g
(11) (d) The insect will crawl up the bowl till the component of
its weight along the bowl is balanced by limiting friction
so, resolving weight perpendicular to the bowl and along
the bowl we get
R
y
f
L
q
mg sinq
mg cosq
R = mg sin
q
f
L
 = mg cos q
tanq =
m
=
m
=
1
R
R
f
R
1
1
L
  ;  
m
=
-
1
y R
y
2 2
m
2
y
2
 = R
2
 – y
2   
;  
 
  y =
1
R
2
1
+ m
So, h = R – y = R –
1
R
2
+ m
= R
ú
ú
û
ù
ê
ê
ë
é
+ m
-
) 1 (
1
1
2
DPP/ P 11
33
(12) (a)
L
h
m
Loss in P .E. in reaching the bottom = mgh and gain in K.E.
reaching the bottom = 
1
2
mv
2
where v is velocity gained by the body in reaching the
bottom
\ Net loss in energy = mgh – 
1
2
 mv
2
work done against friction = FL
\ mgh – 
1
2
mv
2
 = FL ; v = 
2
(mgh FL)
m
-
(13) (a) Let R be the normal reaction on the block exerted by the
floor. The limiting (maximum) force of static friction is
f
s
 = m
s
R = m
s
mg
  = 0.4 × 2kg × 9.8 ms
–2 
= 7.84 N
The applied force F is 2.5 N, that is less than the limiting
frictional force. Hence under the force F, the block does
not move. So long the block does not move, the (adjustable)
frictional force is always equal to the applied force. Thus
the frictional force is 2.5 N.
(14) (b) When the block does not slip on the table surface, it
performs simple harmonic motion along with the table.
x = a sin wt
The instantaneous acceleration of the block is
2
2
dx
dt
 = –w
2
a sin wt
The maximum acceleration is 
2
2
max
dx
dt
 = w
2
a
The maximum force on the block is f
max
 = mw
2
a
where m is its mass. The frictional force on the block is
mmg. since the block is at rest with respect to the table, we
have mw
2
a = mmg
(2pf)
2
a = mg
 Þ a = 
22
g
4f
m
p
 = 
22
0.72 10
4 (3.14) 3
´
´´
 = 0.02 m
(15) (c) Stopping distance is independent on mass.
(16) (a) (i) coefficient of static friction is always greater than the
coefficient of kinetic friction
(ii) limiting friction is always greater than the kinetic friction
(iii) limiting friction is never less than static friction
(17) (d) The system can not remain in equilibrium
(18) (a) (i) In the force applied v/s friction graph : The graph is
a straight line of slope 45º for small F and a straight line
parallel to the F-axis for large F .
(ii) There is small kink on the graph
(19) (a) (i) force of friction between two bodies may be equal to
zero
(ii) bodies may be rough
(20) (b) It is easier to pull a body than to push, because the
friction force is more in pushing than that in pulling
(21) (a)ma = µmg
a = µg
(22) (a)
(1) Kinetic friction is lesser than limiting friction.
(2) In rolling the surfaces at contact do not rub each other.
(3) If a body is at rest and no pulling force is acting on it,
force of friction on it is zero.
(23) (a)
(1) Force of friction is partically independent of microscopic
area of surface in contact and relative velocity between
them. (if it is not high)
(2) Normally with increase in smoothness friction
decreases. But if the surface area are made too smooth by
polishing and cleaning the bonding force of adhesion will
increase and so the friction will increase resulting in 'Cold
welding'
(3) Friction is a non conservative force, i.e. work done
against friction is path dependent.
(24) (c)
(2) Friction may opposes the motion
(4) If the applied force is increased the force of static friction
also increases upto limiting friction.
(25) (a), (26) (a), (27) (a).
F
max
 = kx + µ mg
F
min
 = kx – µ mg
\  F
max
 + F
min
 = 2µ mg
or  2 = 2 µ 10
\  µ = 0.1
F
max
 + F
min
 = 2kx ......... (1)
From graph, F
max
 + F
min
 = 5 and x = 0.1
Putting in eq. (1)
t = 2k (0.1) ; k = 25 N/m
When x = 0.03
kx = 25 × 0.03 = 0.75 N, which is less than µ mg
        = 0.1 × 10 = 1N
\ The block will be at rest, without applying force F.
(28) (b) It is easier to pull a heavy object than to push it on a
level ground. Statement-1 is true. This is because the
normal reaction in the case of pulling is less as compared
b y pus hing. (f = m N). Therefore the functional force is
small in case of pulling.
Statement-2 is true but is not the correct explanation of
statement-1.
(29) (c) W = (force) × (displacement of point of application)
(30) ( d) Statement – 2 is false because friction force may be
more than applied force when body is retarding and
external force is acting on body .