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300J of work is done in slinding a 2 kg ...

300J of work is done in slinding a 2 kg block up an inclined plane of height 10m. Taking g =`10m//s^(2)` , work done against friction is

A

1000 J

B

200 J

C

100 J

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
`(3)`
Loss in potential energy = mgh
`= 2 xx 10 xx 10 = 200 J`
Gain in kinetic energy = work done against friction = 300 -200 = 100 J
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Knowledge Check

  • 300 J of work is done in slide a 2kg block up an inclined plane of height 10m . Taking g = 10 m//s^(2), work done against friction is

    A
    `200 J`
    B
    `100 J`
    C
    zero
    D
    `1000 J`

  • 300 J of work is done in sliding a 2 kg block up on inclined plane of height 10m. Work done against frictin is ( g=10ms^(-2) )

    A
    1000 j
    B
    200 J
    C
    100 J
    D
    Zero

  • If 250 J of work is done in sliding a 5 kg block up an inclined plane of height 4 m. Work done against friction is (g=10ms^(-2))

    A
    `50J`
    B
    `100J`
    C
    `200J`
    D
    Zero
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