KKMishra's Tutorials: Solutions to Problems on "Newton's Laws of Motion"-'H C Verma's Concepts of Physics, Part-I, Chapter-5', EXERCISES-(Problem Number 1 to 12)

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For links to

other chapters - See bottom of the page

EXERCISES (1 to 12)

1. A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.

Diagram for Q 1

Answer: First of all we need to find the acceleration of the block. Here time t=2 s, distance s=10 m, initial velocity u= 0 m/s, acceleration a=? using the formula s=ut+½at² , we get

10=0+½a.4 → 2a=10 → a=5 m/s²

Given that mass m=2 kg, From the relation F=ma, we get,

F = 2 kg* 5 m/s² =10 N.

2. A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it?

Answer: To know the force we need to know both mass and acceleration. Mass m=2000 kg (given) but acceleration (-ve here) has to be calculated from the given data from the equation of kinematics v²=u²+2as.

v=0 m/s, s=4 m, u=40 km/h = 40000m/3600s =100/9 m/s,

So, 0=(100/9)²+2a.4 → a=-10000/(81*8) =-1250/81 m/s²

Now force F=ma =2000 * -1250/81 N = -2500000/81 N =-3.1x104 N (Negative signs of force and acceleration show that their direction is opposite to the direction of the initial velocity.)

3. In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5x10⁶ m/s in travelling one centimeter. Assuming a straight-line motion, find the constant force exerted on the electron. The mass of the electron is 9.1x10^-31 kg.

Answer: Initial velocity u= 0 m/s, Final velocity v= 5x10⁶ m/s, Distance traveled s= 1 cm = 0.01 m, acceleration a =?

using v²=u²+2as → (5x10⁶)² = 0²+2a*0.01 →0.02a=25x10¹²

→ a=12.5x10¹⁴ m/s²

Given that mass m= 9.1x10^-31 kg

Now Force F=ma = 9.1x10^-31 kg * 12.5x10¹⁴ m/s²

→ F=113.75x 10^-17 N ≈1.1x10^-15 N

4. A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s².

Answer: Let us consider the lower block. Force due to gravity which is weight =mg =0.3 x 10 N =3 N. To balance it tension in the string = 3 N.

Now consider the upper block and string. Weight of upper block= 0.2 x 10 N =2 N. Pull by lower string = 3 N, Both downwards. So to balance it the string will pull them upwards with a force =2 N+3 N =5 N. Hence tension in the upper string =5 N.

The picture below will explain it further.

Diagram for problem number 4

5. Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.

Answer: The force F will pull both of the blocks due to the connected string. So the total mass on which F acts, M =m +m =2m.

Diagram for Q - 5

Let the acceleration of the blocks = a.

Since, force =mass*acceleration, the acceleration =force/mass.

→a =F/2m.

Both the blocks will move with acceleration F/2m. Now consider only the rear block and draw a free-body diagram showing the forces acting on it. We see that the only force on it is due to the tension in the connecting string T. We have here the tension force T producing an acceleration a =F/2m in the rear block.

So T =mass*acceleration

= m*(F/2m)

= F/2.

So the tension in the string = F/2.

6. A particle of mass 50 g moves in a straight line. The variation of speed with time is shown in the figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds.

The figure for problem number 6

Answer: Mass of the particle = 50 g = 0.05 kg.

To know the force acting on the particle, we need to know the acceleration of the particle. In the velocity-time graph the slope of the curve gives the instantaneous acceleration. Let us find it for the given instant.

At t=2 seconds

The graph is a straight line with a positive slope. It means the particle has a constant acceleration with magnitude =15/3 =5 m/s².

So the force acting on it = mass x acceleration = 0.05 kg x 5 m/s²

= 0.25 N, it acts along the motion because it is positive.

At t=4 seconds

The graph is horizontal to the time axis meaning thereby the velocity is constant and no acceleration. It can be understood in this way too that the slope which represents acceleration is zero. Since there is no acceleration at t=4 s. so there is no force acting on the particle at this instant, Force= zero.

At t=6 seconds

The graph shows that velocity is uniformly decreasing with time and the acceleration which is represented by the slope is negative. From the graph value of acceleration = -15/3 =-5 m/s².

Force = mass x acceleration = 0.05 x -5 N =-0.25 N

So the force acting on the particle is 0.25 N and the negative sign shows that its direction is opposite to the motion.

7. Two blocks A and B of mass m_A and m_B respectively are kept in contact on a friction-less table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A?

Answer: Both the blocks accelerate simultaneously with the same acceleration, say a. Since block A exerts a force F on block B which mass is m_B. So a=F/m_B.
Diagram for Q -7

Now the force exerted by the experimenter (say P) accelerates both the block's combined mass which is m_A + m_B. So P= (m_A + m_B)*F/m_B = F(1+m_A /m_B).

8. Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.

Answer: First of all we need to know the acceleration 'a' which is negative here. The data given : distance s = 1 mm =1x10^-3 m, initial velocity u= 30 m/s, final velocity v=0 , we use the equation

v²=u²+2as

0²=30²+2a*1x10^-3

→ a=-900/2x10^-3 = -450000 m/s² (Negative sign indicates that it is a retardation)

mass m=4 mg = 4x10^-6 kg

So the force exerted by each drop on the head = ma

= 4x10^-6 kg * 450000 m/s² = 1.80 N.

9. A particle of mass 0.3 kg is subjected to a force F =-kx with k=15 N/m. What will be the initial acceleration if it is released from a point x=20 cm?

Answer: From the given data we have force,

F = -15 N/m*0.20 m

= -3 N,

The negative sign shows that the direction of the force is opposite to the displacement x.

Mass m =0.3 kg.

From the relation F =ma, we get the initial acceleration

a =F/m = -3/0.3 m/s² = -10 m/s²

The negative sign is for the direction of the acceleration that is opposite to the displacement.

So the magnitude of the initial acceleration = 10 m/s².

10. Both the springs shown in the figure (5-E2) are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration?

The figure for problem number 10

Answer: We know the mass of the particle =m. To know the initial acceleration we need to know the net force acting on the block due to the springs.

Since at the original position the springs are unstretched, thus whether we displace the block to the left or to the right, both the springs will apply force along the same direction ie towards the original position opposing the displacement. Hence both the forces by the springs will add and its value =k₁ x + k₂ x = (k₁ + k₂ )x

So at this position, the initial acceleration of the block will be

= Force/mass = (k₁ + k₂ )x /m and its direction will be opposite to the displacement.

11. A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially, block B is near the right end of block A (figure 5-E3). A constant horizontal force of 10 N is applied to block A. All the surfaces are assumed frictionless. Find the time elapsed before the block B separates from A.

The figure for problem number 11

Answer: Since all the surfaces are frictionless so block B will not move but block A will slide under it. When the block A slides its full length 20 cm, the block B will separate from it.

Mass m of block A = 5 kg, Applied force F=10 N ,so acceleration a=F/m

=10/5 m/s² = 2 m/s²

For block A initial velocity u= 0, acceleration a= 2m/s², distance to be traveled s=20 cm =0.20 m, we need to know the time t =? Here we will use the relation s=ut+½at²

0.20 = 0 + ½ *2 t² → t² = 0.2 → t = 0.45 s.

12. A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in the figure (5-E4). Show that the force (assumed equal for both the friends) exerted by each friend on the rope increases as the man moves up. Find the force when the man is at a depth h.

The figure for problem number 12

Answer: Let the mass of the fallen man be m. His weight =mg (downwards)

Let the force exerted by the friends = F = tension in the rope

If the rope makes an angle θ with the vertical near the man, the component of tension in the rope that helps him to lift upwards

= F cos θ; for both the ropes = 2F cos θ

To lift the man it should be more than his weight, At the balancing position, 2F cos θ =mg

→ F=mg/2cos θ .........(i)

It is clear from this equation that the force F is inversely proportional to cos θ. As the man comes up the angle between the rope and vertical = θ increases because the rope tends to be horizontal. But as the θ increases the value of cos θ decreases. And as cos θ decreases the value of F increases. So the force exerted by the friends increases as the man moves up.

From the figure cos θ = h/√{h²+(d/2)²} , putting this value in (i) we get,

F=mg√{h²+(d/2)²}/2h = mg√(4h²+d²)/4h .... It is the value of force exerted by each friend at depth h.

===<<<O>>>===

Links for the chapter -

HC Verma's Concepts of Physics, Chapter-6, Friction,

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→ Newton's laws of motion - Objective - I

Click here for → Newton's Laws of Motion - Objective -II

Click here for → Newton's Laws of Motion - Exercises(13 to 27)

Click here for→ Newton's Laws of Motion-Exercises(Q.No.28 to 42)

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HC Verma's Concepts of Physics, Chapter-4, The Forces

The Forces-

"Questions for short Answers"

Click here for "The Forces" - OBJECTIVE-I

Click here for "The Forces" - Exercises

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Monday, November 2, 2015

Solutions to Problems on "Newton's Laws of Motion"-'H C Verma's Concepts of Physics, Part-I, Chapter-5', EXERCISES-(Problem Number 1 to 12)

My Channel on YouTube ---> SimplePhysics with KK

For links to

other chapters - See bottom of the page

EXERCISES (1 to 12)

1. A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.

Diagram for Q 1

Answer: First of all we need to find the acceleration of the block. Here time t=2 s, distance s=10 m, initial velocity u= 0 m/s, acceleration a=? using the formula s=ut+½at² , we get

10=0+½a.4 → 2a=10 → a=5 m/s²

Given that mass m=2 kg, From the relation F=ma, we get,

F = 2 kg* 5 m/s² =10 N.

2. A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it?

Answer: To know the force we need to know both mass and acceleration. Mass m=2000 kg (given) but acceleration (-ve here) has to be calculated from the given data from the equation of kinematics v²=u²+2as.

v=0 m/s, s=4 m, u=40 km/h = 40000m/3600s =100/9 m/s,

So, 0=(100/9)²+2a.4 → a=-10000/(81*8) =-1250/81 m/s²

Now force F=ma =2000 * -1250/81 N = -2500000/81 N =-3.1x104 N (Negative signs of force and acceleration show that their direction is opposite to the direction of the initial velocity.)

3. In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5x106 m/s in travelling one centimeter. Assuming a straight-line motion, find the constant force exerted on the electron. The mass of the electron is 9.1x10-31 kg.

Answer: Initial velocity u= 0 m/s, Final velocity v= 5x106 m/s, Distance traveled s= 1 cm = 0.01 m, acceleration a =?

using v²=u²+2as → (5x106)² = 0²+2a*0.01 →0.02a=25x1012

→ a=12.5x1014 m/s²

Given that mass m= 9.1x10-31 kg

Now Force F=ma = 9.1x10-31 kg * 12.5x1014 m/s²

→ F=113.75x 10-17 N ≈1.1x10-15 N

4. A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s².

Answer: Let us consider the lower block. Force due to gravity which is weight =mg =0.3 x 10 N =3 N. To balance it tension in the string = 3 N.

Now consider the upper block and string. Weight of upper block= 0.2 x 10 N =2 N. Pull by lower string = 3 N, Both downwards. So to balance it the string will pull them upwards with a force =2 N+3 N =5 N. Hence tension in the upper string =5 N.

The picture below will explain it further. Diagram for problem number 4

5. Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.

Answer: The force F will pull both of the blocks due to the connected string. So the total mass on which F acts, M =m +m =2m.

Diagram for Q - 5

Let the acceleration of the blocks = a.

Since, force =mass*acceleration, the acceleration =force/mass.

→a =F/2m.

So T =mass*acceleration

= m*(F/2m)

= F/2.

So the tension in the string = F/2.

6. A particle of mass 50 g moves in a straight line. The variation of speed with time is shown in the figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds. The figure for problem number 6

Answer: Mass of the particle = 50 g = 0.05 kg.

To know the force acting on the particle, we need to know the acceleration of the particle. In the velocity-time graph the slope of the curve gives the instantaneous acceleration. Let us find it for the given instant.

At t=2 seconds

The graph is a straight line with a positive slope. It means the particle has a constant acceleration with magnitude =15/3 =5 m/s².

So the force acting on it = mass x acceleration = 0.05 kg x 5 m/s²

= 0.25 N, it acts along the motion because it is positive.

At t=4 seconds

At t=6 seconds

The graph shows that velocity is uniformly decreasing with time and the acceleration which is represented by the slope is negative. From the graph value of acceleration = -15/3 =-5 m/s².

Force = mass x acceleration = 0.05 x -5 N =-0.25 N

So the force acting on the particle is 0.25 N and the negative sign shows that its direction is opposite to the motion.

7. Two blocks A and B of mass mA and mB respectively are kept in contact on a friction-less table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A?

Answer: Both the blocks accelerate simultaneously with the same acceleration, say a. Since block A exerts a force F on block B which mass is mB. So a=F/mB.Diagram for Q -7

Now the force exerted by the experimenter (say P) accelerates both the block's combined mass which is mA + mB. So P= (mA + mB)*F/mB = F(1+mA /mB).

8. Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.

Answer: First of all we need to know the acceleration 'a' which is negative here. The data given : distance s = 1 mm =1x10-3 m, initial velocity u= 30 m/s, final velocity v=0 , we use the equation

v²=u²+2as

0²=30²+2a*1x10-3

→ a=-900/2x10-3 = -450000 m/s² (Negative sign indicates that it is a retardation)

mass m=4 mg = 4x10-6 kg

So the force exerted by each drop on the head = ma

= 4x10-6 kg * 450000 m/s² = 1.80 N.

9. A particle of mass 0.3 kg is subjected to a force F =-kx with k=15 N/m. What will be the initial acceleration if it is released from a point x=20 cm?

Answer: From the given data we have force,

F = -15 N/m*0.20 m

= -3 N,

The negative sign shows that the direction of the force is opposite to the displacement x.

Mass m =0.3 kg.

From the relation F =ma, we get the initial acceleration

a =F/m = -3/0.3 m/s² = -10 m/s²

The negative sign is for the direction of the acceleration that is opposite to the displacement.

So the magnitude of the initial acceleration = 10 m/s².

10. Both the springs shown in the figure (5-E2) are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration? The figure for problem number 10

Answer: We know the mass of the particle =m. To know the initial acceleration we need to know the net force acting on the block due to the springs.

So at this position, the initial acceleration of the block will be

= Force/mass = (k1 + k2 )x /m and its direction will be opposite to the displacement.

Answer: Since all the surfaces are frictionless so block B will not move but block A will slide under it. When the block A slides its full length 20 cm, the block B will separate from it.

Mass m of block A = 5 kg, Applied force F=10 N ,so acceleration a=F/m

=10/5 m/s² = 2 m/s²

For block A initial velocity u= 0, acceleration a= 2m/s², distance to be traveled s=20 cm =0.20 m, we need to know the time t =? Here we will use the relation s=ut+½at²

0.20 = 0 + ½ *2 t² → t² = 0.2 → t = 0.45 s.

3. In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5x10⁶ m/s in travelling one centimeter. Assuming a straight-line motion, find the constant force exerted on the electron. The mass of the electron is 9.1x10^-31 kg.

Answer: Initial velocity u= 0 m/s, Final velocity v= 5x10⁶ m/s, Distance traveled s= 1 cm = 0.01 m, acceleration a =?

using v²=u²+2as → (5x10⁶)² = 0²+2a*0.01 →0.02a=25x10¹²

→ a=12.5x10¹⁴ m/s²

Given that mass m= 9.1x10^-31 kg

Now Force F=ma = 9.1x10^-31 kg * 12.5x10¹⁴ m/s²

→ F=113.75x 10^-17 N ≈1.1x10^-15 N

The picture below will explain it further.

Diagram for problem number 4

6. A particle of mass 50 g moves in a straight line. The variation of speed with time is shown in the figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds.

The figure for problem number 6

7. Two blocks A and B of mass m_A and m_B respectively are kept in contact on a friction-less table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A?

Answer: Both the blocks accelerate simultaneously with the same acceleration, say a. Since block A exerts a force F on block B which mass is m_B. So a=F/m_B.
Diagram for Q -7

Now the force exerted by the experimenter (say P) accelerates both the block's combined mass which is m_A + m_B. So P= (m_A + m_B)*F/m_B = F(1+m_A /m_B).

Answer: First of all we need to know the acceleration 'a' which is negative here. The data given : distance s = 1 mm =1x10^-3 m, initial velocity u= 30 m/s, final velocity v=0 , we use the equation

0²=30²+2a*1x10^-3

→ a=-900/2x10^-3 = -450000 m/s² (Negative sign indicates that it is a retardation)

mass m=4 mg = 4x10^-6 kg

= 4x10^-6 kg * 450000 m/s² = 1.80 N.

10. Both the springs shown in the figure (5-E2) are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration?

The figure for problem number 10

= Force/mass = (k₁ + k₂ )x /m and its direction will be opposite to the displacement.