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Standard XII

Physics

Acceleration

A block stops...

Question

A block stops at a distance $0 m$ when it moves with velocity $20 m / s$ on a frictional surface. The value of coefficient of friction is ( $g = 10 m / s^{2})$

0.5

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Solution

The correct option is A $0.5$
Retardation produced by friction = $μ g$
As friction stops the body after moving $80 m$
apply third equation of motion
$v^{2} = u^{2} + 2 a s$
$0 = 400 - 2 \times μ g \times 40$
$μ = 0.5$

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A block stops at a distance 0 m when it moves with velocity 20 m/s on a frictional surface. The value of coefficient of friction is (g=10m/s2)

The correct option is A 0.5Retardation produced by friction =μgAs friction stops the body after moving 80mapply third equation of motionv2=u2+2as0=400−2×μg×40μ=0.5

A block stops at a distance $0 m$ when it moves with velocity $20 m / s$ on a frictional surface. The value of coefficient of friction is ( $g = 10 m / s^{2})$

The correct option is A $0.5$
Retardation produced by friction = $μ g$
As friction stops the body after moving $80 m$
apply third equation of motion
$v^{2} = u^{2} + 2 a s$
$0 = 400 - 2 \times μ g \times 40$
$μ = 0.5$