9303753286127016 A body is moving along a rough horizontal surface with an initial velocity of 10ms-1. If the body comes to rest after travelling a distance of 12m, then the coefficient of sliding friction will be

A body is moving along a rough horizontal surface with an initial velocity of 10ms-1. If the body comes to rest after travelling a distance of 12m, then the coefficient of sliding friction will be

 Sliding friction as the resistance created by any two objects when sliding against each other. This friction is also known as kinetic friction and is defined as the force that is needed to keep a surface sliding along another surface.

The equation for sliding force includes the coefficient of sliding friction times the normal force.

FS = μSFn

Where,

FS = force of sliding friction

μS = Coefficient of sliding friction

Fn = normal force

Solution

Given

u=10ms−1

s=12m

v=0

Find out

We have to determine the coefficient of sliding friction 

Solution

By third equation of motion,
v2=u2−2as

On substituting the known values we get

0=(10)2−2×a×12

0=100 -24a

100=24a

a= 100/24

a= 4.17ms2

Coefficient of sliding friction is given by the formula above

 FS = μSFn

μ=F/ FS 

= a/g

4.17/10 

= 0.417

Answer

The coefficient of friction = 0.417

Balkishan Agrawal

At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s vision for the school is centered on providing a nurturing environment where every student can thrive, learn, and grow.

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