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A man of mass 60kg sitting on ice pushes...

A man of mass 60kg sitting on ice pushes a block of mass of 12kg on ice horizontally with a speed of `5ms^(-1)`.The coefficient of friction between the man and ice and between block and ice is 0.2. If `g =10 ms^(-2)`, the distance between man and the block, when they come to rest is

A

6m

B

6.5m

C

3m

D

7m

Text Solution

Verified by Experts

The correct Answer is:
B
`S_(1)=(v^(2))/(2mug)`, according to law of conservation of momentum, find `v^(1),S_(2)=(v^(1^(2)))/(2mug), S=S_(1)+S_(2)`
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Knowledge Check

  • A man of mass 60kg sitting on ice pushes a block of mass 12kg on ice horizontally with a speed of 5ms^(-1) The coefficient of friction between the man and ice and between block and ice is 0.2 If g =10ms^(2) the distance beteen man and the block when they come to rest is .

    A
    `6m`
    B
    `6.5m`
    C
    `3m`
    D
    `7m`

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    A
    `360m`
    B
    `10m`
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    C
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    D
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