Laws of Motion Test

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Laws of Motion Test - 6

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Question 1
4 / -1

An iron block of sides 50 cm x 8 cm x 15 cm has to be pushed along the lloor. The force required will be niinimum_when the surface in contact with ground is

A
8 cm x 15 cm surface

B
50 cm x 15 cm surface

C
8 cm x 50 cm surface

D
force is same for all surfaces

Solution

Force will be same for all the surfaces.
Question 2
4 / -1

Figure shows a man of mass 55 kg standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s^-1. The net force acting on the man is

A
35 N

B
45 N

C
55 N

D
65 N

Solution

Here,
Mass of the man, M = 55 kg
As the man is standing stationary w.r.t. the belt,
Acceleration of man = Acceleration of belt
Acceleration of man a = 1ms⁻²
Net force on the man,
F = Ma = (55 kg) × (1 m s^-2) = 55N
Force on man is 55N
Question 3
4 / -1

A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m s^-2. The total mass of the crew and passengers is 500 kg. Choose the correct statements from the following. (Take g = 10 m s^-2)
(i) The force on the floor of the helicopter by the crew and passengers is 1.25 x 10⁴ N vertically downwards.
(ii) The action of the rotor of the helicopter on the surrounding air is 6.25 x 10⁴ vertically downwards.
(iii) Theforceon the helicopter due to the surrounding air is 6.25 x 10⁴N vertically upwards.

A
(i) and (ii)

B
(ii) and (iii)

C
(i) and (iii)

D
All the three

Solution

Here,
Mass of hehcopter, M = 2000 kg
Mass of the crew and passengers, m = 500 kg
Vertically upwards acceleration, a = 15 ms^-2
g = 10 m s^-2
(i) Force on the floor by the crew and passengers
= m (g + a) = 500 kg (10 + 15) m s^-2
= 12500 N = 1.25 x 10⁴ N
It acts vertically downwards.
(ii) Action of the rotor of the hehcopter on the surrounding air = (M + m) (g + a)
= (2000 + 500) kg (10 + 15) m s ^-2
= 62500 N = 6.25 x 10⁴ N
It acts vertically downwards
(iii) Force on the helicopter due to the surrounding air is equal and opposite to the action of the rotor of the helicopter on the surrounding air.
Force on the helicopter due to the surrounding air
= 6.25 x 10⁴N
It acts vertically upwards.
Question 4
4 / -1

A person in an elevator accelerating upwards with an acceleration of 2 m s^-2, tosses a coin vertically upwards with a speed of 20 m s^-1. After how much time will the coin fall back into his hand? (Take g = 10 m s^-2)

A
5/3s

B
3/10s

C
10/3s

D
3/5s

Solution

Here, v = 20m s⁻¹, a = 2m s⁻², g = 10m s⁻²
The coin will fall back into the person’s hand after t s.
∴ t =
= 10/3s
Question 5
4 / -1

A person of mass 50 kg stands on a weighing scale on a lift. If the lift is ascending upwards with a uniform acceleration o f 9 m s^-2, what would be the reading of the weighing scale? (Take g = 10m s^-2)

A
50 kg

B
60 kg

C
95 kg

D
100 kg

Solution

The reading on the scale is a measure of the force on the floor by the person. By the Newton’s third law this is equal and opposite to the normal force N on the person by the floor.
∴ When the lift is descending downward with a acceleration of a 9 ms⁻², then
N - 50 × 10 = 50 × 9
or N = 50 × 10 + 50 × 9 = 50(10 + 9)
= 950 N
∴ The reading of weighing machine is 95kg.
Question 6
4 / -1

Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° as shown in the figure. A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.

A
25 N

B
50 N

C
75 N

D
100 N

Solution

As the system is in equilibrium,
∴ T = W …(i)
and T = mg sin30^∘ …(ii)
From (i) and (ii), we get

W = mg sin 30^∘ = (100 N) × 1/2 = 50 N
Question 7
4 / -1

Two blocks of masses 10 kg and 20 kg are connected by a massless string and are placed on a smooth horizontal surface as shown in the figure. If a force F = 600 N is applied to 10 kg block, then the tension in the string is

A
100 N

B
200 N

C
300 N

D
400 N

Solution

Here, m(1) = 10kg, m = 20kg, F = 600N
Let T be tension of the string and a be common acceleration of the system.
a =
= 20 ms^-2
When force F is applied on 10kg block, then the tension in the string is
T₂ = m₂a = (20kg)(20ms−²) = 400N
Question 8
4 / -1

In the question number 91, if a force F is applied to 20 kg block, then the tension in the string is

A
100 N

B
200 N

C
300 N

D
400 N

Solution

When a force F is applied on 20kg block, then the tension in string is
T = m₂a = (10kg)(20ms^-2) = 200N
Question 9
4 / -1

Two masses of 5 kg and 3 kg are suspended with the help of massless inextensible strings as shown in figure. The whole system is going upwards with an acceleration of 2 m s^-2. The tensions T( and T₂ are respectively (Takeg = 10 m s^-2)

A
96 N, 36 N

B
36 N, 96 N

C
96 N, 96 N

D
36 N, 36 N

Solution

The free body diagram of 3 kg block is as shown in the fig. (a).
The equation of motion of 3 kg block is T₂ − 3g = 3a
T₂ = 3(a + g) = 3(2 + 10) = 36N ............(i)
The free body diagram of 5 kg is as shown in the Fig.(b).

The equation of motion of 5kg block is
T₁ −T₂ − 5g = 5a 1
= 5(a + g) + T₂
= 5(2 + 10) + 36 = 96N (Using (i))
Question 10
4 / -1

Two blocks each of mass M are resting on a frictionless inclined plane as shown in figure. Then

A
The block A moves down the plane.

B
The block B moves down the plane.

C
Both the blocks remain at rest.

D
Both the blocks move down the plane

Solution

The force acting down the plane for blocks A and B are:

Since F_A > F_B the block A moves down the plane.
Question 11
4 / -1

In the system shown in the figure, the acceleration of 1 kg mass is

A
g/4 downwards

B
g/2 downwards

C
g/2 upwards

D
g/4 upwards

Solution

If a is downward acceleration of 4 kg block, the upward acceleration of 1 kg block must be 2a.

If T is tension in each part of string, then equations of motion of the two blocks are
4a = 4g − 2T ...(i)
1 × 2a = T −1 g ..(ii)
or 4a = 2T −2g ...(iii)
Adding (i) and (iii), we get
8a = 2g, a = g / 4
∴ Acceleration of 1 kg block = 2a = g/2 upwards.
Question 12
4 / -1

Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is

A
g/4

B
g/5

C
g/8

D
g/6

Solution

Let a be the common acceleration of the system and T be tension of the string. The equations of motion of two blocks are
T − 8g = 8a …(i)
and 12g −T = 12a

Adding (i) and (ii), we get
4g = 20a or a = g/5
Question 13
4 / -1

A monkey of mass 40 kg climbs on a massless rope which can stand a maximum tension of 500 N. In which of the following cases will the rope break? (Take g = 10 m s^-2)

A
The monkey climbs up with an acceleration of 5 m s^-2.

B
'Hie monkey climbs down with an acceleration of 5 m s^-2.

C
The monkey climbs up with a uniform speed of 5 m s^-1.

D
The monkey falls down the rope freely under gravity.

Solution

Here, mass of monkey, m = 40kg
Maximum tension the rope can stand, T = 500N
Tension in the rope will be equal to apparent weight of the monkey (R).
The rope will break when R exceeds T.(a) When the monkey climbs up with an acceleration a = 5m s⁻²
R = m (g + a) = 40(10 + 5) = 600N
∴ R > T
Hence, the rope will break.
(b) When the monkey climbs down with an acceleration a=5m s⁻²
R = m (g − a) = 40(10 − 5) = 200N
∴ R < T
Hence, the rope will not break.
(c) When the monkey climbs up with a uniform speed v = 5m s⁻¹, its acceleration a = 0
∴ R = mg = 40 × 10 = 400N
∴ R < T
Hence, the rope will not break.
(d) When the monkey falls down the rope freely under gravity
a = g
∴ R = m(g − a) = m(g − g) = zero
Hence, the rope will not break.
Question 14
4 / -1

Two blocks of masses of 40 kg and 30 kg connected by a weightless string passing over a frictionless pulley as shown in the figure.

The acceleration of the system would be

A
0.7 m s^-2

B
0.8 m s^-2

C
0.6 m s^-2

D
0.5 m s^-2

Solution

Here, m₁ = 40kg

m₂ = 30kg, θ = 30^∘
Let T be the tension in the string and a be the acceleration of the system.
Their equations of motion are m₁gsin30^∘ −T = m₁a …(i)
T − m₂gsin30^∘ = m₂a …(ii)
Adding (i) and (ii), we get
(m₁ + m₂) a = (m₁ − m₂) gsin30^∘
Substituting the given values, we get
(40 + 30) a = (40 − 30) × 9.8 × 1/2 = 49
a = 49/70 = 0.7m s⁻²
Question 15
4 / -1

A book is lying on the tab le. What is the angle between the action of the book on the table and the reaction of the table on the book?

A
0^o

B
45^o

C
90^o

D
180^o

Solution

The weight (W) and the action (A) both will act in same direction and parallel to each other. So the angle between them should be zero.