# 71 - Newton's laws of motion and friction Questions Answers

Hi please slove these question also...

A ) **Two particles of mass 3 kg and 4 kg are connected by a light inelastic string passing over a smooth fixed pulley. The system is released from rest with the string taut and both particles at a height of 2 m above the ground. Find the velocity of the 3 kg mass when the 4 kg mass reaches the ground, and find when the 4 kg mass reaches the ground. **

**B) Briefly describe an experiment to find the coefficient of friction between brick and tile. You may assume access to a tile slab, a number of bricks and basic scientific equipment, such as a pulley, weighing machine, etc.**

Please reply ASAP...Waiting for your reply.

**Asked By: RAAJ**

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**Joshi sir comment**

(A)4g-T = 4a

T-3g = 3a

solve for a

use h = 0t+1/2at^{2} for getting t

and v = 0 +at for v

(B) get F applied and acceleration of the body

then use F-kmg = ma

for getting k (coefficient of friction)

Please slove the below mention question:

A) A uniform ladder of weight w rests on rough horizontal ground against a smooth vertical wall. The vertical plane containing the ladder is perpendicular to the wall and the ladder is inclined at an angle θ to the *vertical. *Prove that, if the ladder is on the point of slipping and µ is the coefficient of friction between it and the ground, then tanθ = 2µ

B) Two particles of mass 3 kg and 4 kg are connected by a light inelastic string passing over a smooth fixed pulley. The system is released from rest with the string taut and both particles at a height of 2 m above the ground. Find the velocity of the 3 kg mass when the 4 kg mass reaches the ground, and find when the 4 kg mass reaches the ground.

C) A box of mass 14 kg is placed in the back of a van. The coefficient of friction between the box and the floor is 0.5. What happens to the box if the lorry moves off with an acceleration of

(a) 4 ms^{-2}

(b) 5 ms^{-2}

(c) 8 ms^{-2}

(Take g = 10 ms^{-2})

Can i get answer before 3pm today. Please i request you to help me out from the issue...

**Asked By: RAAJ**

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**Joshi sir comment**

(a)let R and N are reactions by the ground and wall then for equilibrium

μR = N

R = w

and about ground point

N*lcosθ = w*lsinθ/2

now solve

(b) 4g-T = 4a

T-3g = 3a

solve for a

use h = 0t+1/2at^{2} for getting t

and v = 0 +at for v

(c) for a = 8 pseudo force = 14*8 and friction max = 0.5*14*10 so box will fall in backward direction

A 50 kg gymnast falls freely from a height of 4 m on to a trampoline. The trampoline then bounces her back upward with a speed equal to the speed at which she first struck the trampoline. What is the average force the trampoline applies on the gymnast ?

A) 50 N B) 200 N C) 500 N D) 2000 N E) more information is required

**Asked By: VAIBHAV GUPTA**

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**Joshi sir comment**

a package of mass 10kg is released from rest on a rough slope inclined at 25 degrees to the horizontal. after 2 seconds the package has moved 4m down the slope. find the coefficient of friction between the package and the slope.

a box of mass 2kg is pushed up a rough plane by a horizontal force of magnitude 25N. the plane is inclined to the horizontal at an angle of 10 degree. given that the coefficient of friction between the box and the plane is 0.3, find the acceleration of the box.

**Asked By: AMAR KHAN**

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**Joshi sir comment**

first part is correctly explained by Sarika

for second part make components of forces as 25 N and weight along and perpendicular to the incline, friction μN will be along downward incline and N will be the resultant of components of weight and 25 N perpendicular to the plane.

In order to raise a mass of 100 kg, a man of mass 60 kg fastens a rope to it and passes the rope over a smooth pulley. He climbes the rope with an acceleration 5g/4 relative to the rope. The tension of the rope is:

(a) 1432 N

(b) 928 N

(c) 1218 N

(d) 642 N

**Asked By: RAJIV**

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**Joshi sir comment**

eq. for upward motion of 100 Kg is T-100g = 100a (1)

for man eq. is T-60g = 60[(5g/4)-a] from ground frame

solve?

sir

there is a statement ie

FRICTION IS SELF ADJUSTABLE , is it right or not ?? The problem is only static frictn is self adjustable , so wat shld be the ans for only frictn ??

**Asked By: SARIKA**

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**Joshi sir comment**

in a general way the statement is correct

__If a particle is moving on a circular path with constant speed, then the angle between the direction of acceleration and its position vector wrt centre of circle will be__

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

it should be 180 degree

__Two blocks of masses 2 kg and 4 kg are hanging with the help of massless string passing over an ideal pulley inside an elevator. The elevator is moving upward with an acceleration g/2. The tension in the string connected between the blocks will be__

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

equation 1

4g+4g/2-T = 4a

equation 2

T-2g-2g/2 = 2a

now solve

Sir, can third law of Newton be derived from second law??

actually as we all know that second law is real law so first & second law can be derived, but on the other side it is written that they are THE LAWS ARE INDEPENDENT THUS CAN'T BE DERIVED FROM ANY OTHER LAW (acc to HC Verma)

sir sometime ans is yes while sometime no , in different books different ans , i consulted this query to many other teachers also but some favours while some teachers not , so what you suggest?

is this depends upon questn setting person

**Asked By: SARIKA**

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**Joshi sir comment**

THE LAWS ARE INDEPENDENT

A body is moving on a circle of radius 80m with a speed 20m/s which is decreasing at the rate of 5m/s^{2} at an istant. What is the angle made by its acceleration with its velocity?

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

You should write that calculate

What is the angle made by its acceleration with its velocity at the instant for which conditions are given.

solution

centripetal acceleration = 20*20/80 = 5 m/s^{2}

tangential acceleration = 5 m/s^{2}

so angle = tan^{-1}(5/5) = 45^{0}