Force and Pressure Test

Result

Force and Pressure Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

A spring scale reads 20 newtons as it pulls a 1 kg mass across a table. What is the magnitude of the force exerted by the mass on the spring scale?

A
49 N

B
20 N

C
15 N

D
4 N

Solution

According to Newton's 3^rd law of motion, to every action, there is an equal and opposite reaction. So, magnitude of the force exerted by the mass on the spring scale is 20 N.
Question 2
1 / -0

Three blocks of masses, i.e. M₁ = 1 kg, M₂ = 2 kg and M₃ = 3 kg are in contact with each other on a frictionless horizontal surface as shown in the figure. What is the horizontal force 'F' needed to push the blocks as one unit with an acceleration of 2 ms^–2?

A
6 N

B
4 N

C
12 N

D
2 N

Solution

a =
2 =
12 N = F
Question 3
1 / -0

A force acts for 5 s on a stationary body of mass 200 kg after which the force ceases to act. The body moves through a distance of 20 m in the next 2 s. What is the magnitude of force applied?

A
800 N

B
200 N

C
100 N

D
400 N

Solution

t = 5 sec
m = 200 kg
Since the body moves 20 m in 2 sec after the application of force, distance of 10 m is covered in 1 sec. So, the object moves with the velocity of 10 m/s.
F = 200 x = 400 N
Question 4
1 / -0

A force acts for 4 s on a body of mass 3.2 kg initially at rest. The force then ceases to act and the body moves through a distance of 4 m in the next second. What is the magnitude of the force?

A
3.2 N

B
1.6 N

C
12.8 N

D
6.4 N

Solution

t = 4 s
m = 3.2 kg
u = 0
Since the body covers a distance of 4 m after applying the force, the velocity (v) will be 4 m/s.
F = ma
F = 3.2 × = 3.2 N
Question 5
1 / -0

An object of mass 300 g is resting on a frictionless table. Calculate the acceleration of the object when it is acted upon by a force of 0.03 N.

A
1 ms^–2

B
10 ms^–2

C
3 ms^–2

D
0.1 ms^–2

Solution

F = m*a
a = F/m = 0.03/0.3
= 0.1 m/s^-2
Question 6
1 / -0

A force of 20 N acts on a body of mass 4 kg for 6 s initially at rest. What is the velocity acquired by the body and the change in momentum of the body?

A
15 m/s, 30 kg m/s

B
15 m/s, 60 kg m/s

C
30 m/s, 120 kg m/s

D
30 m/s, 80 kg m/s

Solution

F = m.a
F = m(v-u)/t
v = (20*6)/4 = 30m/s
Change in momentum = m* (v-u) = 4* 30 = 120 kg m/s
Question 7
1 / -0

A car of mass 500 kg moving at a speed of 108 km/h is stopped in 20 s. What is the force applied by the brakes?

A
2700 N

B
1500 N

C
750 N

D
900 N

Solution

M = 500 kg
u (initial velocity) = 108 km/h
v (final velocity) = 0m/s
T = 20 s u = 108 x = 30 m/s
F = ma = m x
F = 750 N is applied by brakes
Question 8
1 / -0

A car is moving with a uniform velocity 60 ms^-1. It comes to rest in 4 s by applying a force of 3000 N. What is the change in the momentum of the car?

A
3000 kg m/s

B
1200 kg m/s

C
1500 kg m/s

D
12000 kg m/s

Solution

u = 60 m/s
t = 4 sec
F = 3000 N
F = ma = m x
3000 =
P = 3000 x 4
= 12000 kg x m/s
Question 9
1 / -0

Two people push a car for 4.8 s with a combined net force of 100 N. What is the impulse provided to the car?

A
480 Ns

B
0.048 Ns

C
20.8 Ns

D
2.08 Ns

Solution

F=
100 =
P =
P = 480 Ns
Question 10
1 / -0

An object A of mass 3 kg is moving with a velocity of 3 ms^–1 and collides head on with an object B of mass 2 kg moving in opposite direction with a velocity of
2 ms^–1. After the collision, both the objects stick and move with a common velocity. What is the velocity of the objects after the collision?

A
2.6 ms^–1

B
5 ms^–1

C
1 ms^–1

D
2 ms^–1

Solution

m_A = 3 kg, u_A = 3 m/s; m_B = 2kg, u_B = 2m/s - Before collision

Acc. to the law of conservation of momentum:
m_A u_A+ m_B u_B = (m_A + m_B) V
3 x 3 + 2 x (-2) = 5 x V (since objects are moving opposite to each other)
9-4 = 5v
5 = 5V
V = 1 m/s
Question 11
1 / -0

Impulse provided when 1 N force is applied for 1 s on a body is equal to

A
1 kg

B
1 kg

C
1 kg

D
1 kg

Solution

F =
m x a =
m x
(Since v= , t = s)
m = kg
Kg x
P = 1 kg
Question 12
1 / -0

Force of friction on an object lying on a table is 29.4 N. What is the mass of the object if the coefficient of friction between the object and the table surface is 0.3?

A
2.94 kg

B
10 kg

C
98 kg

D
8.82 kg

Solution

F (Force of friction) = 29.4 N
(coefficient of friction) = 0.3
F = (R = mg)
29.4 = 0.3 x m x g
29.4 = 0.3 x m x 9.8
m = 10 kg
Question 13
1 / -0

A bullet of mass 50 g is horizontally fired with a velocity of 200 ms^–1 from a pistol of mass 4 kg. What is the magnitude of recoil velocity of the pistol?

A
2.5 ms^–1

B
2 ms^–1

C
3 ms^–1

D
1.5 ms^–1

Solution

m_B = 50 g = 0.05 kg
v_B = 200 m/s
m_p = 4 kg
Initial momentum = final momentum
m_B v_B + m_p V_p= 0
0.05 x 200 + 4 x v_p = 0
-= v_p
- 2.5 m/s = v_pThe magnitude of recoil velocity = 2.5 m/s
Question 14
1 / -0

Two balls A and B of masses 2 kg and 3 kg are moving with velocities 3 ms^–1 and 2 ms^–1 respectively, towards each other. During the collision, they stick together. What will be the velocity of the combined balls after the collision?

A
2.4 ms^–1

B
0 ms^–1

C
5 ms^–1

D
1 ms^–1

Solution

m_A = 2 kg, m_B = 3kg
u_A = 3 m/s; u_B = 2 m/s
Acc. to the law of conservation of momentum:
m_A u_A + m_B u_B = (m_A+ m_B) v
2 x 3 + (-(3 x 2)) = (2 + 3) v (since balls are moving towards each other)
6 – 6 = 5 x v
0 m/s = v
Question 15
1 / -0

A ball of mass 200 g is moving with a velocity of 6 ms^–1. It is brought to rest after travelling a distance of 2 m by applying a resistive force. What is the value of the resistive force applied?

A
1.2 N

B
0.6 N

C
1.8 N

D
2 N

Solution

m = 200 g = 0.2 kg
U = 6 m/s
S = 2 m
Using v² – u² = 2aS
0 – 36 = 2 x F/m x S
- 36 = 2 x F/0.2 x 2
- 36 =
- 36 = N
Resistive force (F) = 1.8 N
Question 16
1 / -0

An applied force of 24 N [forward] causes a steel block to move across a horizontal greased steel surface. What is the approximate value of mass of the block?
(Take coefficient of friction = 0.15) (g = 9.8 m/s²)

A
2.4 kg

B
36 kg

C
1.6 kg

D
16 kg

Solution

F= 24 N
u = 0.15
F = uR
24 = 0.15 x m x 9.8
m = 16.3 kg
Question 17
1 / -0

A force causes an acceleration of 15 ms^–2 in a body of mass 300 g. What acceleration will be caused by the same force on a body of mass 5 kg?

A
900 ms^–2

B
0.9 ms^–2

C
15 ms^–2

D
1.8 ms^–2

Solution

a = 15m/s², m = 300 g = 0.3 kg
F = ma = 0.3 x 15 = 4.5 N
Using F = ma,
m = 5 kg
4.5 = 5 x a
a = ²
Question 18
1 / -0

A soccer player heads the ball with an average net force of 12 N for 0.12 s. What is the impulse provided to the soccer ball?

A
0.01 Ns

B
1.44 Ns

C
100 Ns

D
12 Ns

Solution

F=
12 =
P = 12 x =
= 1.44 Ns
Question 19
1 / -0

The following figure shows the velocity-time graph of a particle of mass 500 g moving in a straight line. What is the force acting on the particle?

A
10 N

B
1.25 N

C
1250 N

D
3.25 N

Solution

F = ma = m x (v-u)/(t₂ - t₁)
U = 0; v = 20 m/s
t₂ = 8 sec, m = 500g = 0.5 kg
t₁ = 0 sec
F =
=
F = 1.25 N
Question 20
1 / -0

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass 'm' as shown in the figure. A force F is applied to one end of the rope. What is the force that the rope exerts on the block?

A
F

B

C

D

Solution

Acceleration of the combined system (a) = F/m+M
Force that rope exerts on the block = M x F/m+m