Home
Class 11
PHYSICS
A block of mass 1 kg is stationary with ...

A block of mass `1 kg` is stationary with respect to a conveyer belt that is accelerating with `1 m//s^(2)` upwards at an angle of `30^(@)` as shown in figure. Which of the following is`//`are correct?

A

Force of friction on block is 6 N upwards

B

Force of friction on block is `1.5` N upwards

C

Contact force between the block and belt is `10.5 N`

D

Contact force between the block and the belt is `5 sqrt3N`

Text Solution

Verified by Experts

The correct Answer is:
A, C

Block is moving upwards due to friction `f_(r) - mg sin 30^(@) = ma`
` (##GRB_AM_PHY_C07_E01_104_S01.png" width="80%">
`implies f_(r) - 1 xx 10 xx (1)/(2) = 1 xx l implies f_(r) = 6N`
Contact force is the resultant of N and `f_(r)`
`= sqrt(N^(2) + f_(r)^(2)) = sqrt((mg cos 30)^(2) + (6)^(2))`
`=10.5 N`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • FRICTION AND CIRCULAR MOTION

    GRB PUBLICATION|Exercise Assertion - Reason|12 Videos
  • FRICTION AND CIRCULAR MOTION

    GRB PUBLICATION|Exercise Matrix - Match|5 Videos
  • FRICTION AND CIRCULAR MOTION

    GRB PUBLICATION|Exercise OBJECTIVE QUESTIONS ( only one choice is correct )|80 Videos
  • FORCE AND NEWTONS LAWS OF MOTION

    GRB PUBLICATION|Exercise Comprehension|12 Videos
  • MOTION IN A STRAIGHT LINE

    GRB PUBLICATION|Exercise Comprehension type Queston|14 Videos

Similar Questions

Explore conceptually related problems

A block of mass 1 kg is stationary with respect to a conveyor belt that is accelerating with 1 m//s ^(2) upwards at an angle of 30^(@) as shown in figure. Contact force between the block & belt is 3.5x. What is the value of x ?

Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms^-2 What is the net on the man.

Knowledge Check

  • A block of mass 1 kg is stationary with respect to a converyor belt that is accelerationg with 1 m//s^(2) upward at an angle of 30^(@) as shown in figure . Which of the following statement (s) is /are correct ? (g=10m//s^(2))

    A
    Force of friction on the block is 1 N upwards
    B
    Force of friction on the block is 1.52 N upwards
    C
    Contact force between the block and the belt is 10.5 N
    D
    Contact force between the block and belt is `5sqrt(3) `N
  • Figure shows a man of mass 55 kg standing stationary with respect to a horizontal conveyor belt that is accelerating with 1m s^(-2) . The net force acting on the man is

    A
    35N
    B
    45N
    C
    55N
    D
    65N
  • Figure shows a man standing stationary with respect to a horizontal conveyer belt that is accelerating with 1 ms^(-2) What is the net force on the man? (Mass of the man = 65 kg)

    A
    35N
    B
    45N
    C
    55N
    D
    65N
  • Similar Questions

    Explore conceptually related problems

    Figure shown a man standing stationary with respect to a horizontal converyor belt that is accelerationg with 1m//s^(-2) . What is the net force on the man?If the coefficient of ststic friction between the man's shoes and the belt is 0.2 upto what maximum acceleration of the belt can the man continue to be stationary relative to the belt? Mass of the man = 65kg (g =9.8m//s^(2))

    A block of mass m is stationary with respect to a rough wedge as shown in figure. Starting from rest in time t, (m=1 kg, theta=30^(@) , a=2ms^(-2) , t=4s) work done on block

    A block of mass m is stationary with respect to a rough wedge as shown in figure. Starting from rest in time t, (m = 1 kg, theta=30^(@), a = 2 m//s^(2) , t = 4 s) work done on block :

    A man of mass 65kg is standing stationary with respect to a conveyor belt which is accelerating with 1m//s^(2) . if mu_(s) is 0.2 the net force on the man and the maximum acceleration of the belt so that the man is stationary relative to the belt are (g =10 m//s^(2)) .

    A wooden block of mass 1kg and density 800 Kg m^(-3) is held stationery, with the help of a string, in a container filled with water of density 1000 kg m^(-3) as shown in the figure. If the container is moved upwards with an acceleration of 2 m s^(-2) , then the tension in the string will be ( take g=10ms^(-2) )