Home
Class 12
PHYSICS
A wedge is moving rightwards on which a ...


A wedge is moving rightwards on which a block of mass 10 kg is placed on it. Friction coefficient between the eedge and the block is 0.8. Select correct alternative (s) among the following options. [take `g=10(m)/(s^(2))`]

A

if wedge is moving with constant velocity then friction acting on block is 64N.

B

if wedge is moving with cosntant velocity then acceleration of block is zero.

C

if wedge is moving with `veca=2(veci)(m)/(s^(2))` then friction acting on block is 44N.

D

if wedge is moving with `veca=10(veci)(m)/(s^(2))` then friction is 20N. Downward on the block along the inclined.

Text Solution

Verified by Experts

The correct Answer is:
B, C, D

(A,B) if moving with constant velocity then `a=0`
so friction available `=mumgcostheta`
`=(0.8)(10)(10)((4)/(5))=64N`.
But `mgsintheta=60N`
So required friction is 60 N.
So net force is zero.
`(C)a=2hat(i)f=mgsin37^(@)-macos37^(@)=44N`
`(D)f=mgsin37^(@)-macos37^(@)=-20N`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • TEST PAPERS

    RESONANCE|Exercise PT-01|30 Videos

  • TEST PAPERS

    RESONANCE|Exercise Pt-02|30 Videos

  • TEST PAPERS

    RESONANCE|Exercise PART - II PHYSICS|107 Videos

  • SIMPLE HARMONIC MOTION

    RESONANCE|Exercise Advanced Level Problems|13 Videos

  • TEST SERIES

    RESONANCE|Exercise PHYSICS|127 Videos

Similar Questions

Explore conceptually related problems

If coeffiecient of friction between the block of mass 2kg and table is 0.4 then out acceleration of the system and tension in the string. (g=10m//s^(2))

A force of 200N is applied as shown in the figure on block of 3 kg. The coefficient of friction between the block and the wall is 0.3. Find the friction acting on the block. (in N) {Take g = 10 m//s^(2)

Knowledge Check

  • A wedge is moving rightwards on which a block of mass 10kg is placed on it. Friction coefficient between the wedge and the block is 0.8.[ takes g=10m//s^(2)] . Select correct alternative (s) among the following options.

    A
    If wedge is moving with constant velocity then friction acting on block is `64N`.
    B
    If wedge is moving with constant velocity then acceleration of block is zero.
    C
    If wedge is moving with `vec(a)=2(hat(i))m//s^(2)` then friction acting on block is `44N`
    D
    If wedge is moving with `vec(a)=10(hat(i))m//s^(2)` then friction is `20N`, downward on the wedge along the inclined.

  • A block of mass 1 kg is placed on a wedge shown in figure. Find out minimum coefficient of friction between wedge and block to stop the block on it.

    A
    0.6
    B
    0.9
    C
    1
    D
    0.2

  • A force of 100N is applied on a block of mass 3 kg as shown in the figure-2.164. The coefficient of friction between the surface and the block is 0.25.The frictional force acting on the block is: (Take g=10m/s^(2) )

    A
    15 N downwards
    B
    25 N upwards
    C
    20 N downwards
    D
    30 N upwards

    • Similar Questions

    Explore conceptually related problems

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

    A vehicle is moving on a road with an acceleration a=20 m//s^(2) as shown in figure. The frictional coefficient between the block of mass (m) and the vehicle so that block is does not fall downward is (g=10m//s^(2))

    A force F is applied on a block of mass 5 kg as shown in figure. Co-efficient of friction between the wall and clock is (sqrt(3))/(2). (g = 10m//s^(2)) Select correct option(s)

    A block of mass 50 kg slides over a horizontal distance of 1m. If the coefficient of friction between their surface is 0.2, then work done against friction is (take g = 9.8 m//s^(2) ) :

    A block of mass "5kg" is placed on a wedge having inclination of 37^(@) with the horizontal. Coefficient of friction between block and wedge is 0.8. Then select the correct statement.

    Home
    Profile