Force Test

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Force Test - 8

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Question 1
1 / -0

A particle is in straight line motion with uniform velocity. A force is not required

A
to increase the speed

B
to decrease the speed

C
to keep the same speed

D
to change the direction

Solution

Force acting on an object moving in a straight line will change its speed. Hence, to maintain the same speed, no force should be applied.
Question 2
1 / -0

An object will continue accelerating until

A
resultant force on it begins to decrease

B
its velocity changes direction

C
the resultant force on it is zero

D
the resultant force is at right angles to its direction of motion

Solution

From the relation = Ma, we know that if sum of all the forces on a body is non-zero, the body will have an acceleration. Hence, a body will continue accelerating until the resultant force on it is zero.
Question 3
1 / -0

A boy sitting on the top most berth in the compartment of a train, which is just going to stop on a railway station, drops an apple aiming at the open hand of his brother sitting vertically below him at a distance of about 2m. The apple will fall

A
in the hand of his brother

B
slightly away from the hands of his brother in the direction of motion of the train

C
slightly away from the hands of his brother in the direction opposite to the direction of motion of the train

D
none of the above

Solution

As the train is about to stop (de-accelerating) then, the tendency for a freely falling body inside the train will be to move slightly in the direction of motion of the train. This is because of inertia of motion. Hence, the apple will fall slightly away from the hands of his brother in a direction of motion of the train.
Question 4
1 / -0

A man of weight W is standing on a lift, which is moving upward with acceleration `a`. The apparent weight of the man is

A
W(1 + a/g)

B
W

C
W(1 - a/g)

D
W (1 - a²/g²)

Solution

Lets consider Newton`s law ( = ma) acting on the person. The overall acceleration of the person is in an upward direction. So, `ma` is positive. The only external forces acting on the person are force of gravity acting in a downward direction (-W = - mg) and the normal force `N` acting in an upward direction.
So, = - mg + N = ma
N = mg + ma (1)
Let `` be the weight of the person
W = mg (2)
From (1) and (2),
N = W + ma
= W
= W
N = W
Question 5
1 / -0

A lift is moving upward with an acceleration of 3 m/s² and g = 9.8 m/s². To carry a bag of mass 5 kg in his hands, a man in the lift has to exert a force of

A
15 N

B
49 N

C
34 N

D
64 N

Solution

Lift is moving up with a = 3 m/s²
Acceleration due to gravity (g) = 9.8 m/s²

F - mg = ma
F = mg + ma
= 5(9.8) + 5(3)
= 49.0 + 15
= 64 N
Question 6
1 / -0

Gravel is dropped into a conveyer belt at a rate of 0.5 kg/s. The extra force required in Newton to keep the belt moving at 2 m/s is

A
1

B
2

C
4

D
0.5

Solution

Rate at which the gravel is dropped = 0.5 kg/s
Velocity with which the belt is moving = 2 m/ s
Force required = 0.5 2 = 1 N
Question 7
1 / -0

A ball falls from rest from a height h onto a floor, and rebounds to a height h/4. The coefficient of restitution between the ball and the floor is

A

B

C

D

Solution

C.O.R. =
Here, H =h (Initial height)
h` = (Final height after rebound)
C.O.R. =
Question 8
1 / -0

A bullet weighing 50 g leaves the gun with a velocity of 30 m/s. If the recoil speed imparted to the gun is 1 m/s, the mass of the gun is

A
15 kg

B
30 kg

C
1.5 kg

D
20 kg

Solution

Now, from the law of conservation of momentum we know that
Initial momentum = Final momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Here, m₁ = mass of gun, v₁ = recoil velocity of gun, u₁= initial velocity of gun
m₂ = mass of bullet, v₂ = recoil velocity of bullet, u₂= initial velocity of bullet
0 = m(-1) +
M = 1.5 kg
Question 9
1 / -0

Two balls, each of mass 0.25 kg, are moving with 3m/s and 1m/s towards each other in a straight line collide. The balls stick together after the collision. The magnitude of the final velocity of the combined mass is

A
4 m/s

B
2 m/s

C
1 m/s

D
(1/2) m/s

Solution

Let `v` be the final velocity of system. Now, according to the "law of conservation of momentum",
Initial momentum = Final momentum
(0.25 3) + (0.25 -1) = (0.25 + 0.25) v
0.75 - 0.25 = 0.5 v
0.5 = 0.5 v
1 = v
v = 1 m/ s
Question 10
1 / -0

A body of mass m moving with a constant velocity v hits another body of the same mass moving with same velocity v but in the opposite direction and sticks to it. The velocity of the compound body, after collision, is

A
v

B
2v

C
zero

D
v/2

Solution

mv) + (-mv) = (2m) (v_f)
0 = (2m) V_f
V_f = 0
Question 11
1 / -0

Two blocks of masses m₁ and m₂ are placed in contact with each other on a horizontal platform. The coefficient of friction between the platform and the two blocks is the same. The platform moves with an acceleration. The force of interaction between the blocks is

A
zero in all cases

B
zero only if m₁ = m₂

C
non-zero only if m₁> m₂

D
non-zero only if m₁ < m₂

Solution

Two blocks of masses m₁ and m₂ are placed in contact with each other on a horizontal platform. The coefficient of friction between the platform and the two blocks is the same. The platform moves with an acceleration. The force of interaction between the blocks is zero in all cases.
Question 12
1 / -0

A car C of mass m₁ rests on a plank P of mass m₂. The plank rests on a smooth floor. The string and the pulley are ideal. The car starts and moves towards the pulley with acceleration. Which of the following assertions is correct?

A
If m₁ > m₂, the string will become slack.

B
If m₁ < m₂, the string will remain under tension.

C
If m₁ = m₂, the string will have no tension, and C and P will have acceleration of equal magnitude.

D
C and P will have accelerations of different magnitudes if m₁ m₂.

Solution

If m₁ = m₂, then the strings will have no tension and the car and the plank will have equal acceleration because the plank is placed on another floor.
Question 13
1 / -0

In the figure given below, two masses m and m` are tied with a thread passing over a pulley. m` is on a frictionless horizontal surface. If acceleration due to gravity is g, the acceleration of m` in this arrangement will be

A
g

B
mg/(m + m`)

C
mg/m`

D
mg/(m - m`)

Solution

Force balance on mass m:

mg - T = ma …….. (1)
Force balance on mass m`:
T = m`a …….. (2)
Eliminating T from (1) and (2), we get
mg - m`a = ma
= a
Question 14
1 / -0

A spring of force constant k is cut into two pieces such that one piece is double the length of the other. What will be the force constant of the longer piece?

A
(2/3) k

B
(3/2) k

C
3k

D
6k

Solution
Question 15
1 / -0

Two bodies A and B each of mass M are fixed together by a mass less spring. A force F acts on the mass B as shown in figure. At the instant, the mass A has acceleration a. What is the acceleration of mass B?

A
(F/ M) - a

B
a

C
- a

D
(F/M)

Solution

Let tension in spring = T
Then, T = Ma (where `a` is acceleration of body `A`)
Let a` be acceleration of body B.
Force balance on B,
F - T = M a`
F - Ma = Ma`
a` =
Question 16
1 / -0

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor is (taking g = 10 m/s²)

A
2.8 N

B
8 N

C
2 N

D
Zero

Solution

M = 2 kg
g = 10 m/s²
Normal force = N = mg = 20 N
Force of static friction = mg = 0.4 2 10 = 8 N
If a force which is less than 8 N is applied, then the block will not move. Hence, frictional force of the same magnitude as that of the force applied, acts in an opposite direction.
Hence = f = 2.8 N
Question 17
1 / -0

A line-man of mass 60 kg is holding a vertical pole. The coefficient of static friction between his hands and the pole is 0.5. If he is able to climb up the pole, what should be the minimum force with which he will press the pole with his hands? (g = 10 m/s²)

A
1200 N

B
600 N

C
300 N

D
150 N

Solution

Mass of the line-man, m = 60 kg
Gravity, g = 10 m/s²
Coefficient of static friction, μ_s = 0.5
Weight of the line-man,

Let the minimum force with which the line-man should press the pole with his hands in order to climb up the pole be F.
Then,
Question 18
1 / -0

A block of mass 2 kg rests on a rough horizontal plank. The coefficient of friction between the plank and the block is 0.2. If the plank is pulled horizontally with a constant acceleration of 4 m/s², the distance moved by the block on the plank in 5 second starting from rest (in m) is (Take g = 10 m/s²)

A
5

B
10

C
25

D
50

Solution

Given data:
m = 2 kg
μ = 0.2
Pulling acceleration, a₁ = 4 m/s²
Gravity, g = 10 m/sNet acceleration of the block,

As the block is initially at rest, initially velocity of the block u = 0 m/s
Distance moved by the block on the plank in 5 seconds,
Question 19
1 / -0

Two men of unequal masses hold on to the two sections of a light rope passing over a smooth light pulley. Which of the following is/are possible?

A
The lighter man is stationary while the heavier man slides with some acceleration.

B
The heavier man is stationary while the lighter man slides with some acceleration.

C
The two men slide with the same acceleration in the same direction.

D
The two men slide with different acceleration in opposite directions.

Solution

Let mass of the lighter person = m
T = mg
Let mass of the heavier person = M
Mg - T = Ma
Mg - mg = Ma
So, the lighter man is stationary while the heavier man slides with some acceleration.
Question 20
1 / -0

A force-time graph for the motion of a body is shown in the figure. Change in linear momentum between 0 and 8 s, is

A
zero

B
4 Ns

C
8 Ns

D
None of these

Solution

Change in momentum = Area under F-t curve
Total area under graph = 0 (+ve and -ve areas are equal.)