Home
Class 11
PHYSICS
A body moving on the surface of the eart...

A body moving on the surface of the earth at 14 `"m.s"^(-1)` comes to rest due to friction after covering 50 m. Find the coefficient of friction between the body and the earth's surface. Given, acceleration due to gravity = 9.8 `"m.s"^(-2)`.

Text Solution

Verified by Experts

Initial velocity, u = 14 `"m.s"^(-1)`, final velocity, v = 0, displacement , s = 50 m and mass of the body = m. If retardation is a then using the formula `v^(2) = u^(2) - 2` as,
a `= (u^(2))/(2s) = (14xx14)/(2xx50) = 1.96 "m.s"^(-2)`
`:.` Kinetic friction `f_(k)` = ma = 1.96 mN and normal force, R = mg = 9.8m N
`:.` Coefficient of friction,
`mu = (f_(k))/(R) = (1.96m)/(9.8 m)` =0.2.
Promotional Banner

Topper's Solved these Questions

  • FRICTION

    CHHAYA PUBLICATION|Exercise Section Related Questions|9 Videos

  • FRICTION

    CHHAYA PUBLICATION|Exercise HIGHER ORDER THINKING SKILL ( HOTS ) QUESTIONS|14 Videos

  • FIRST AND SECOND LAW OF THERMODYNAMICS

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|18 Videos

  • HYDROSTATICS

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|1 Videos

Similar Questions

Explore conceptually related problems

A body moving on ground with a velocity of 14 m//s comes to rest due to friction after moving ove a distance of 50 m. Detennine the co-efficient of friction between the body and the ground.

A body of mass 2 kg is pulled with a velocity 2 m cdot s^(-1) on horizontal surface. What will be the heat produced in 5s, if the coefficient of friction between the body and the surface is 0.2? Given, J = 4.2 J cdot cal^(-1) , g = 9.8 m cdot s^(-2) .

A tram is moving with an acceleration of 49 cm .s^(-2) using 50% of its motor power the remaining 50% is used up to overcome friction. Find the coefficient of friction between the wheel and the tram line.

The acceleration due to gravity on the surface of the earth is 9.8 m*s^(-2) . The size of a planet is the same as that of the earth, but its density is twice the density of the earth. The value of the acceleration due to gravity on that planet is

A body is rolling on the ground with a velocity of 1 m/s. After travelling a distance of 5m, it comes to rest. Find the coefficient of friction.Take g = 10 m/s^2 .

A car of mass 500 kg is moving up along an inclined surface of slope 1/(25) at a constant speed of 72 km*h^(-1) . If the coefficient of friction between the road and the car wheel is 0.1 , find the power of the car engine (g= 9.8 m*s^(-2) ).

What is the escape velocity of a meteorite situated 1800 km above the surface of the earth? Given, the radius of the earth =6300 km and the acceleration due to gravity on the earth's surface =9.8m*s^(-2) .