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A 40kg slab rests on a frictionless floo...

A 40kg slab rests on a frictionless floor as shown in the figure. A 10kg block rests on the top of the slab. The static coefficient of friction between the block and slab is `0.60` while the kinetic friction is `0.04` . The 10kg block is acted upon by a horizontal force 100N. if `g=9.8m//s^(2)` , the resulting acceleration of the slab will be.

A

`1m//s^(2)`

B

`1.5m//s^(2)`

C

`2m//s^(2)`

D

`6m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Limiting friction between block and slab `=mu_(s)m_(A)g`
`=0.6xx10xx10=60N`
But applied force on block `A` is `100N` . So the block will slip over a slab.
Now kinetic friction works between block and slab`Fk=mu_(k)m_(A)g=0.4xx10xx10=40N`
This kinetic friction helps to move the slab
:. Acceleration of slab `=(40)/(m_(B))=(40)/(40)=1m//s^(2)` .
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Knowledge Check

  • A 40 kg slab rests on a frictionless floor as shown in the figure. A 10 kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force 100 N. If g = 9.8 m/s^(2) , the resulting acceleration of the slab will be

    A
    `1 m//s^(2)`
    B
    `1.5 m//s^(2)`
    C
    `2 m//s^(2)`
    D
    `6 m//s^(2)`

  • A 40 kg slb rest on a frictionless floor as shown in the figure. A 10 kg block rests on the top of the slab. The static corfficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40 the 10 kg block is acted upon by a horizontal force 100N. If g=9.8(m)/(s^2) , the resulting acceleration of the slab will be.

    A
    `1(m)/(s^2)`
    B
    `1.5(m)/(s^2)`
    C
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    D
    `6(m)/(s^2)`

  • A 40 kg slab rests on a frictionless floor . A 10 kg block rests on top of the slab (Fig . 7. 121). The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40 . The 10 kg block is acted upon by a horizontal force of 100 N . if g = 9.8 m//s^(2) the resulting acceleration of the slab will be :

    A
    `0. 98 m//s^(2)`
    B
    `1.47 m//s^(2)`
    C
    `1.52 m//s^(2)`
    D
    `6.1 m // s^(2)`

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