Check here the Latest detailed Answers and Solutions as per the latest guidelines of Page No 113: - Chapter 5 Laws Of Motion class 11 ncert solutions Physics - [2024-2025]

Page No 113: - Chapter 5 Laws Of Motion class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 5.33:

A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s^–2

(b) climbs down with an acceleration of 4 m s^–2

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Fig. 5.20

Answer:

Case (a)

Mass of the monkey, m = 40 kg

Acceleration due to gravity, g = 10 m/s

Maximum tension that the rope can bear, T_max = 600 N

Acceleration of the monkey, a = 6 m/s² upward

Using Newton’s second law of motion, we can write the equation of motion as:

T – mg = ma

∴T = m(g + a)

= 40 (10 + 6)

= 640 N

Since T > T_max, the rope will break in this case.

Case (b)

Acceleration of the monkey, a = 4 m/s² downward

Using Newton’s second law of motion, we can write the equation of motion as:

mg – T = ma

∴T = m (g – a)

= 40(10 – 4)

= 240 N

Since T < T_max, the rope will not break in this case.

Case (c)

The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0.

Using Newton’s second law of motion, we can write the equation of motion as:

T – mg = ma

T – mg = 0

∴T = mg

= 40 × 10

= 400 N

Since T < T_max, the rope will not break in this case.

Case (d)

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g

Using Newton’s second law of motion, we can write the equation of motion as:

mg – T = mg

∴T = m(g – g) = 0

Since T < T_max, the rope will not break in this case.

Question 5.34:

Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B?What happens when the wall is removed? Does the Answer to (b) change, when the bodies are in motion? Ignore the difference between μ_s and μ_k.

Fig. 5.21

Answer:

(a) Mass of body A, m_A = 5 kg

Mass of body B, m_B = 10 kg

Applied force, F = 200 N

Coefficient of friction, μs = 0.15

The force of friction is given by the relation:

fs = μ (m_A + m_B)g

= 0.15 (5 + 10) × 10

= 1.5 × 15 = 22.5 N leftward

Net force acting on the partition = 200 – 22.5 = 177.5 N rightward

As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.

Hence, the reaction of the partition will be 177.5 N, in the leftward direction.

(b) Force of friction on mass A:

f_A = μm_Ag

= 0.15 × 5 × 10 = 7.5 N leftward

Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward

As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., 192.5 N acting leftward.

When the wall is removed, the two bodies will move in the direction of the applied force.

Net force acting on the moving system = 177.5 N

The equation of motion for the system of acceleration a,can be written as:

Net force causing mass A to move:

F_A = m_Aa

= 5 × 11.83 = 59.15 N

Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N

This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion.

Question 5.35:

A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s^–2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.

Answer:

(a) Mass of the block, m = 15 kg

Coefficient of static friction, μ = 0.18

Acceleration of the trolley, a = 0.5 m/s²

As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:

F = ma = 15 × 0.5 = 7.5 N

This force is acted in the direction of motion of the trolley.

Force of static friction between the block and the trolley:

f = μmg

= 0.18 × 15 × 10 = 27 N

The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.

When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.

(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

Question 5.36:

The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s^–2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Fig. 5.22

Answer:

Mass of the box, m = 40 kg

Coefficient of friction, μ = 0.15

Initial velocity, u = 0

Acceleration, a = 2 m/s²

Distance of the box from the end of the truck, s‘ = 5 m

As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:

F = ma

= 40 × 2 = 80 N

As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction f, acting between the box and the floor of the truck. This force is given by:

f = μmg

= 0.15 × 40 × 10 = 60 N

∴Net force acting on the block:

F_net = 80 – 60 = 20 N backward

The backward acceleration produced in the box is given by:

a_back

Using the second equation of motion, time t can be calculated as:

Hence, the box will fall from the truck after from start.

The distance s, travelled by the truck in is given by the relation:

= 20 m

Question 5.37:

A disc revolves with a speed of rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

Answer:

Coin placed at 4 cm from the centre

Mass of each coin = m

Radius of the disc, r = 15 cm = 0.15 m

Frequency of revolution, ν = rev/min

Coefficient of friction, μ = 0.15

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Coin placed at 4 cm:

Radius of revolution, r‘ = 4 cm = 0.04 m

Angular frequency, ω = 2πν

Frictional force, f = μmg = 0.15 × m × 10 = 1.5m N

Centripetal force on the coin:

F_cent.

= 0.49m N

Since f > F_cent, the coin will revolve along with the record.

Coin placed at 14 cm:

Radius,

= 14 cm = 0.14 m

Angular frequency, ω = 3.49 s^–1

Frictional force, f‘ = 1.5m N

Centripetal force is given as:

F_cent.

= m × 0.14 × (3.49)²

= 1.7m N

Since f < F_cent., the coin will slip from the surface of the record.

Question 5.38:

You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?

Answer:

In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

The net force acting on the motorcyclist is the sum of the normal force (F_N) and the force due to gravity (F_g= mg).

The equation of motion for the centripetal acceleration ac, can be written as:

F_net = mac

Normal reaction is provided by the speed of the motorcyclist. At the minimum speed (v_min), F_N = 0

Question 5.39:

A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Answer:

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction, μ = 0.15

Frequency of rotation, ν = 200 rev/min

The necessary centripetal force required for the rotation of the man is provided by the normal force (F_N).

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force (f = μF_N) acting upward.

Hence, the man will not fall until:

mg < f

mg < μF_N = μmrω²

g < μrω²

The minimum angular speed is given as:

Question 5.40:

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for .What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ?Neglect friction.