Home
Class 11
PHYSICS
A block of mass m is on an inclined L pl...

A block of mass m is on an inclined L plane of angle 0. The coefficient of M, friction between the block and the plane is `mu and tantheta gt mu`. The block is e held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from `P_(1)=mg(sin0 -mucostheta)"to" P_(2)= mg(sintheta + mucostheta)`, the frictional force f versus P graph will look like

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
a

Initially the friction force is upward as `P lt mg sin theta` .As `P` increases frictional force decreases and at value of `P= mg sin theta` friction will be zero Afterward friction will the direction and start increasing
Promotional Banner

Topper's Solved these Questions

  • NEWTON'S LAWS OF MOTION 2

    CENGAGE PHYSICS ENGLISH|Exercise Multiple correct answer type|3 Videos
  • NEWTON'S LAWS OF MOTION 2

    CENGAGE PHYSICS ENGLISH|Exercise Assertion -reasoning|2 Videos
  • NEWTON'S LAWS OF MOTION 2

    CENGAGE PHYSICS ENGLISH|Exercise True And False|4 Videos
  • NEWTON'S LAWS OF MOTION 1

    CENGAGE PHYSICS ENGLISH|Exercise Integer|5 Videos
  • PROPERTIES OF SOLIDS AND FLUIDS

    CENGAGE PHYSICS ENGLISH|Exercise INTEGER_TYPE|2 Videos

Similar Questions

Explore conceptually related problems

A block of mass m is on an inclined plane of angle theta . The coefficient of friction between the block and the plane is mu and tanthetagtmu . The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P_1=mg(sintheta-mucostheta) to P_2=mg(sintheta+mucostheta) , the frictional force f versus P graph will look like

A block of mass m sides down an inclined right angled trough .If the coefficient of friction between block and the trough is mu_(k) acceleration of the block down the plane is

A block of mass 2 kg slides down an incline plane of inclination 30°. The coefficient of friction between block and plane is 0.5. The contact force between block and plank is :

A block of mass m slides down an inclined plane of inclination theta with uniform speed. The coefficient of friction between the block and the plane is mu . The contact force between the block and the plane is

A block of mass m slides down an inclined plane which makes an angle theta with the horizontal. The coefficient of friction between the block and the plane is mu . The force exerted by the block on the plane is

A block of mass m is lying on an inclined plane. The coefficient of friction between the plane and the block is mu . The force (F_(1)) required to move the block up the inclined plane will be:-

An ice cube is kept on an inclined plane of angle 30^(@) . The coefficient to kinetic friction between the block and incline plane is the 1//sqrt(3) . What is the acceleration of the block ?

A block of mass m is placed at rest on an inclination theta to the horizontal. If the coefficient of friction between the block and the plane is mu , then the total force the inclined plane exerts on the block is

A block of mass m is at rest on a rough inclined plane of angle of inclination theta . If coefficient of friction between the block and the inclined plane is mu , then the minimum value of force along the plane required to move the block on the plane is

A block of mass m = 2kg is resting on a inclined plane of inclination 30^(@) The coefficient of friction between the block and the plane is mu = 0.5 what minimum force F (in newton) should be applied perpendicular to the plane on the block so that the does not slip on the plane?