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

A body is projected with a velocity of 20m/s and at an angle of 30° with the ground. The change in the direction of its velocity cannot be (during its motion).

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

I think options are necessary for this problem.

but if we analyse we can say that the angle will be between 0 and 120 degree

Q.On a spring balance,

(i)10N force is applied on both the ends

(ii)10N force is applied on one end & no force on the other end.

(iii)10N is applied on one end and 20N is applied at the other.

Calculate the reading of the spring balance in each case.

Please reply fast.

**Asked By: ASUTOSH SAHOO**

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**Joshi sir comment**

i) 10 N

ii) 0 N, and it will be accelerated

iii) 10 N, and it will be accelerated

Sir,

please explain in detail how friction helps in walking?Please involve all the action and reaction forces involved in the process..Also what is the reason that we can't walk on a frictionless surface?

**Asked By: ASUTOSH SAHOO**

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**Joshi sir comment**

in accelerated motion friction will be forward and due to the same reason we take small steps when we walk in ice due to reduce friction. Without friction you can not accelerate or retard on the surface.

is its force dat depends on acceleratn or its acceleratn that depends upon force??

**Asked By: SARIKA**

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**Joshi sir comment**

both are proportional to each other

if a particle is moving on a circular path with constant speed then the angle between the direction of acceleration and its position vector with respect to centre of circle will be?

**Asked By: SHWETA BHARDWAJ**

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**Joshi sir comment**

since speed is constant so only centripetal acc. will be present and it will be parallel to position vector starting from centre of the circle.

1. A mass "m" moving with a velocity "u" hits a surface at an angle "Q" with the normal at the point of hitting . How much force does it exerts, if no energy is lost ?

2. A thin cicular loop of radius "R" rotates about its vertical diameter with angular frequency "W" . show that a small bead on the wire loop remains at its lowermost point for W ≤ (root of g/R) . what is the angle made by the radius vector joining the center to the bead with the vertical downwards direction for W= ( roots of 2g/R) . neglect friction .

3. A rear side of a truck is open and a box of 40 kg is placed 5m away from the open end . the coefficient of friction b/w box and surface in 0.15 on a straight road , the truck starts from rest and acclerates with 2m/s^{2}. at what distance from the starting point does the box fall off the truck .

4. Straight from rest , a mass "m" slides down on inclined plane "Q" in a time "n times " the time to slide down the same lenght in absence of friction . Find the coefficient of friction .

**Asked By: PRIYANKA DAHIYA**

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**Joshi sir comment**

1) angle is measured from vertical so vertical component of velocity = ucosθ

and since no energy is lost so returning velocity in vertical direction will be same

so impulse = (mucosθ) - (-mucosθ)

force cant be calculated without time of impact.

2)

at the position given in diagram mrω^{2}cosθ = mgsinθ for equilibrium

so rω^{2}cosθ = gsinθ and r = Rsinθ

so Rsinθω^{2}cosθ = gsinθ so Rω^{2}cosθ = g or ω = √[g/Rcosθ] so ω ≥ √[g/R]

thus for W ≤ (root of g/R), bead will remain at the lowermost point.

for second part compare W= ( roots of 2g/R) and ω = √[g/Rcosθ]

3) pseudo in opposite direction = ma

frictional force in the direction of motion of truck = 0.15mg

so equation of motion of box in opposite direction is ma - 0.15mg = ma'

so 2-1.5 = a' or a' = 0.5

distance = 5m

so time taken upto fall of box can be obtained by s = 1/2 a' t^{2}

after calculating t, use again s = 1/2 (2) t^{2 }for finding distance travelled by truck.

4) in first case with friction s = 1/2 (gsinθ-μgcosθ) (nt)^{2}

in second case s = 1/2 (gsinθ) t^{2}

now solve.

**Asked By: SOORAJ YADAV**

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**Joshi sir comment**

s_{1} = 0*10+1/2*a*100 = 50a

s_{1}+s_{2 }= 0*20+1/2*a*400 = 200a

now solve

A BALLOON WITH ITS CONTENTS WEIGHING 160N IS MOVING DOWN WITH AN ACCELERATION OF g/2 ms^{-2}. THE MASS TO BE REMOVED FROM IT SO THAT THE BALLOOON MOVES UP WITH AN ACCELERATION g/3 ms^{-2} IS

**Asked By: K TEJASAI**

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**Joshi sir comment**

according to the given conditions

for downward journey 160-R = (160/g)g/2 = 80 so R = 80

now after removing m kg the new equation

R - [160/g - m]g = [160/g - m]g/3

put the value of R and get m

एक लिफ्ट में एक सिक्का लिफ्ट के फिर्श से 2m की उचाई पर छोड़ा जाता है, लिफ्ट की उचाई 10m है | लिफ्ट 11m/s^{2}के त्वरण से नीचे की ओर गति कर रही है | वह समय जिसके पश्चात सिक्का लिफ्ट से टकराएगा ?

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**Joshi sir comment**

the coin will fall with 10 m/s^{2} and acc. of lift is given as 11 m/s^{2}. so coin will move upward with acc. 1 m/s^{2} relative to lift and initial velocity of coin = 0 relative to lift. so it will strike the roof of the lift

use s = ut + 1/2 at^{2}

so 8 = 0 + 1/2 (1) t^{2}

so t = 4

A block of mass 10 kg is released on rough incline plane. Block start decending with acceleration 2 m/s² . Kinetic friction force acting on block is ( take g= 10m/s² )

1. 10N 2. 30N

**Asked By: AVANTIKA PURI**

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**Joshi sir comment**

question is not correct