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Class 11th Laws of Motion

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

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Vishal Baghel

Contributor-Level 10

Mass of the block = 15 kg

Coefficient of static friction between the block and the trolley $μ$ = 0.18

Acceleration of the trolley = 0.5 m/s2

(a) Force experienced by block, F = MA = 15 $*$ 0.5 = 7.5 N. This fore acts in the direction of motion of the trolley

Force of friction, Ff = $μ m g$ = 0.18 $* 15 * 10$ N = 27 N

Force experienced by the block is less than the friction, hence for a stationary observer on the ground, the block will be stationary

(b) When an observer moves with the trolley, the trolley will appear to be at rest

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Class 11th Laws of Motion

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 μs and μk

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Vishal Baghel

Contributor-Level 10

Mass of the body A, $m_{A}$ = 5 kg

Mass of the body B, $m_{B}$ = 10 kg

Applied force = 200 N

Coefficient of friction between the bodies and the table, μs = 0.15

(a) The frictional force is given by the relation

fs = $μ$ ( $m_{A}$ + $m_{B}$ )g = 0.15 $* ((5 + 10)) =$ 22.5 N. towards left

Hence, the force on the partition = 200 – 22.5 N = 177.5 N, acting rightwards

According to Newton's 3rd law, the reaction of the partition will be 177.5 N, acting towards left

(b) Force of friction on mass A is given by

Fa = $μ$ mAg = 0.15 $* 5 * 10 =$ 7.5 N, acting leftward

The net force exerted by mass A on mass B = 200 -7.5 N =

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Mass of the body A, $m_{A}$ = 5 kg

Mass of the body B, $m_{B}$ = 10 kg

Applied force = 200 N

Coefficient of friction between the bodies and the table, μs = 0.15

(a) The frictional force is given by the relation

fs = $μ$ ( $m_{A}$ + $m_{B}$ )g = 0.15 $* ((5 + 10)) =$ 22.5 N. towards left

Hence, the force on the partition = 200 – 22.5 N = 177.5 N, acting rightwards

According to Newton's 3rd law, the reaction of the partition will be 177.5 N, acting towards left

(b) Force of friction on mass A is given by

Fa = $μ$ mAg = 0.15 $* 5 * 10 =$ 7.5 N, acting leftward

The net force exerted by mass A on mass B = 200 -7.5 N = 192.5 N, acting rightward

An equal amount of force will be exerted by mass B on A = 192.5 N, acting leftward

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

The 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 = ( $m_{A}$ + $m_{B}$ )a

a = Net force / ( $m_{A}$ + $m_{B}$ )= 177.5 / (5+10) = 11.833 m/s²

Net force causing mass A to move

Fa = $m_{A}$ a = 5 $* 11.833 = 59.17 N$

Net force exerted by mass A on mass B = 192.5 – 59.17 N = 133.33 N

This force acts in the direction of motion. According to Newton's 3rd law, force exerted by mass B on mass A will be 133.33 N.

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6 months ago

Class 11th Laws of Motion

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

(c) Climbs up with a uniform speed of 5 m s^-1

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

(Ignore the mass of the rope)

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Vishal Baghel

Contributor-Level 10

Mass of the monkey = 40 kg

Maximum tension of the rope Tmax = 600 N

(a) When the monkey climbs up with an acceleration of 6m / s²

Tension in the rope, T – mg = ma

T = m (g+a) = 40 (10+6)= 640 N

Since T > Tmax, the rope will break

(b) When the monkey climbs down with an acceleration of 4m/s²

Tension in the rope T is given by mg – T = ma

T = m (g-a) = 40 (10-4) N = 240 N

Since T < T max, the rope will not break

(c) Climbs up with an uniform speed of 5 m/s

In this case, the acceleration = 0

The tension in the rope is given by

T –mg = ma

T = mg = 40 $* 10 = 400 N$

Since T < Tmax, the rope will not break

(d) When the monkey falls down freely under gravity

The tension in the rope is given by the equat

...more

Mass of the monkey = 40 kg

Maximum tension of the rope Tmax = 600 N

(a) When the monkey climbs up with an acceleration of 6m / s²

Tension in the rope, T – mg = ma

T = m (g+a) = 40 (10+6)= 640 N

Since T > Tmax, the rope will break

(b) When the monkey climbs down with an acceleration of 4m/s²

Tension in the rope T is given by mg – T = ma

T = m (g-a) = 40 (10-4) N = 240 N

Since T < T max, the rope will not break

(c) Climbs up with an uniform speed of 5 m/s

In this case, the acceleration = 0

The tension in the rope is given by

T –mg = ma

T = mg = 40 $* 10 = 400 N$

Since T < Tmax, the rope will not break

(d) When the monkey falls down freely under gravity

The tension in the rope is given by the equation

T + mg = ma, since a = g, T = 0

Since T < Tmax, the rope will not break

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6 months ago

Class 11th Laws of Motion

5.32 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?

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Vishal Baghel

Contributor-Level 10

Mass of the block = 25 kg

Mass of the man = 50 kg

Acceleration due to gravity = 10 m/s2

Weight of the block = 250 N

Weight of the man = 500 N

In the 1st case, the man lifts the block directly, applying an upward force, same as the block = 250 N

Due to Newton's 3rd law of motion, the downward reaction of the man on the floor = 500 N + 250 N = 750 N

In the 2nd case, the man applies a downward force of 25 kg wt. According to Newton's 3rd law, the action on the floor by the man 500N-250N = 250N

The man should adopt the second case

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6 months ago

Class 11th Laws of Motion

5.31 A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg.

What provides the centripetal force required for this purpose — The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ?

0 Follower 15 Views

Vishal Baghel

Contributor-Level 10

Given

Radius, r = 30m

Speed, v = 54 km/h = 15 m/s

The mass of the train, m = 106 kg

(a) The required centripetal force is provided by the rails to the wheels of the train

(b) The angle of banking required to prevent wearing out of the rails

$\tan ? θ$ = $v^{2}$ / rg = $15^{2}$ / (30 $* 10)$

$θ$ = 37 $°$

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6 months ago

Class 11th Laws of Motion

5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?

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Vishal Baghel

Contributor-Level 10

The speed of the aircraft, v = 720 km/h = 200 m/s

The angle of banking = 15 $°$

From the relation $\tan ? θ$ = $v^{2}$ /rg we get r = $v^{2}$ / ( g $* \tan ? θ)$ = 14928 m

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6 months ago

Class 11th Laws of Motion

5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) The force on the 7th coin (counted from the bottom) due to all the coins on its top

(b) The force on the 7th coin by the eighth coin

(c) The reaction of the 6th coin on the 7th coin

0 Follower 62 Views

Vishal Baghel

Contributor-Level 10

(a) The force on the 7th coin is due to the 3 coins kept above it

The weight of the 3 coins = 3m

Force exerted on the 7th coin is (3mg) N and the force is acting vertically downwards

(b) The force on the 7th coin by the 8th coin will be the force exerted by the 3 coins above it = (3mg) N, acting downwards

(c) The 6th coin will experience the force of 4 coins above it, acting downwards = 4mg

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6 months ago

Class 11th Laws of Motion

5.28 A stream of water flowing horizontally with a speed of 15 m s^-1 gushes out of a tube of cross-sectional area 10^-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?

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Vishal Baghel

Contributor-Level 10

Speed of water, v = 15 m/s

Cross-sectional area of the tube, A = 10^-2 m²

Density of water $ρ$ = 1000 kg/m3

Mass of the water hitting the wall = $ρ * A * v$ = 1000 $*$ 10^-2 $* 15$ = 150 kg/s

Force exerted on the wall because of impact of water = mass $*$ velocity = 150 $* 15$ N = 2250 N

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6 months ago

Class 11th Laws of Motion

5.27 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) Force on the floor by the crew and passengers

(b) Action of the rotor of the helicopter on the surrounding air

(c) Force on the helicopter due to the surrounding air

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Vishal Baghel

Contributor-Level 10

Given

Mass of the helicopter m₁ = 1000 kg

Mass of the crew and passenger m₂ = 300 kg

The vertical acceleration of the helicopter, a = 15 m/s² and acceleration due to gravity g = 10 m/s²

The total mass of the system, m = m₁ + m2 = 1000 + 300 = 1300 kg

(a) Force on the floor by the crew and passengers :

R – m₂g =m₂a

R = m₂ (g+a) = 300 $*$ ( 10 + 15 ) N = 7500 N

(b) Action of the rotor of the helicopter on the surrounding air

R' – mg = ma

R' = m (g+a) = 1300 $* (10 + 15) N$ = 32500 N

(c) Force on the helicopter due to the surrounding air

It is the reaction of the force applied by the rotor on the air. As action and reaction are

...more

Given

Mass of the helicopter m₁ = 1000 kg

Mass of the crew and passenger m₂ = 300 kg

The vertical acceleration of the helicopter, a = 15 m/s² and acceleration due to gravity g = 10 m/s²

The total mass of the system, m = m₁ + m2 = 1000 + 300 = 1300 kg

(a) Force on the floor by the crew and passengers :

R – m₂g =m₂a

R = m₂ (g+a) = 300 $*$ ( 10 + 15 ) N = 7500 N

(b) Action of the rotor of the helicopter on the surrounding air

R' – mg = ma

R' = m (g+a) = 1300 $* (10 + 15) N$ = 32500 N

(c) Force on the helicopter due to the surrounding air

It is the reaction of the force applied by the rotor on the air. As action and reaction are equal and opposite (Newton's 3rd law), the force of reaction, F = 32500 N, acting vertically upwards.

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Class 11th Laws of Motion

5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]

Lowest Point Highest Point

(a) mg – T₁ mg + T₂

(b) mg + T₁ &nbs

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Lowest Point Highest Point

(a) mg – T₁ mg + T₂

(b) mg + T₁ mg – T₂

(c) mg +T₁ – (m $v_{1}^{2}$ ) / R mg – T₂+ (m $v_{1}^{2}$ ) / R

(d) mg – T₁ – (m $v_{1}^{2}$ ) / R mg + T₂ + (m $v_{1}^{2}$ ) / R

T₁ and v₁ denote the tension and speed at the lowest point. T₂ and v₂ denote corresponding values at the highest point.

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Vishal Baghel

Contributor-Level 10

The net force at the lowest point is denoted by (mg – T₁)and the net force at the highest point is denoted by (mg + T₂), hence option (a) is correct. The forces mg and T₁are in mutually opposite direction at the lowest point and mg and T₂are in the same direction at the highest point.

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