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Standard XII

Physics

Finding Friction's Magnitude

A block A of ...

Question

A block A of mass 45 kg is placed on another block B of mass 123 kg. Now block B is displaced by external agent by 50 cm horizontally towards right. During the same time block A just reaches to the left end of block B. Initial and final positions are shown in figures. The work done on block A in ground frame is

-18 J

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18 J

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36 J

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-36 J

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Solution

The correct option is C 18 J
As there is relative motion so kinetic friction will act on A in the direction of motion of B (or towards right).

$f = μ m g = 0.2 \times 45 \times 10 = 90 N$

$S = 50 - 30 = 20 c m = 0.2 m$

$∴ W = f S = 18 J$

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The correct option is C 18 JAs there is relative motion so kinetic friction will act on A in the direction of motion of B (or towards right).f=μmg=0.2×45×10=90NS=50−30=20cm=0.2m∴W=fS=18 J

The correct option is C 18 J
As there is relative motion so kinetic friction will act on A in the direction of motion of B (or towards right).

$f = μ m g = 0.2 \times 45 \times 10 = 90 N$

$S = 50 - 30 = 20 c m = 0.2 m$

$∴ W = f S = 18 J$