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An ice block is projected vertically up ...

An ice block is projected vertically up with a velocity 20 `ms^(-1)`. The amount of ice that melt when it reaches the ground and if the loss of P.E. is converted into heat energy if the mass of ice block is 4.2 kg

A

2.5gm

B

2.5kg

C

0.25kg

D

0.25gm

Text Solution

Verified by Experts

The correct Answer is:
A
Maximum height attained, `h=(u^(2))/(2g)`
`W=JHrArrmgh=J(xL_("ice"))`
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Knowledge Check

  • An ice cube of mass M_(0) is given a velocity v_(0) on a round horizontal surface with coefficient of friction mu . The block is at its melting point and latent heat of fusion of ice is L . The block receive heat only due to the friction forces and all work is converted into heat. Find the mass of the remaining ice block after time t.

    A
    `m=m_(0)e^(-(2mug)/(L)(V_(0)t+(1)/(2)mu"gt"^(2)))`
    B
    `m=m_(0)e^(-(2mu)/(L)(V_(0)t-(1)/(2)mu"gt"^(2)))`
    C
    `m=m_(0)e^(-(3mu)/(L)(V_(0)t-(1)/(2)mu"gt"^(2)))`
    D
    `m=m_(0)e^(-(2mu)/(L)(V_(0)t-mu"gt"^(2)))`

  • A 42 kg block of ice moving on rough horizontal surface stops due to friction, after sometime. If the initial velocity of the decelerating block is 4ms^-1 , the mass of ice (in kg) that has melted due to the heat generated by the friction is (Latent heat of ice is 3.36xx10^5Jkg^ -1

    A
    `1xx10^-3`
    B
    `1.5xx10^-3`
    C
    `2xx10^-3`
    D
    `2.5xx10^-3`

  • 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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