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Standard XII

Physics

Problems with Pseudo Force

A weight of 4...

Question

A weight of 4.9 N rests on a rough horizontal floor that has a co-efficient of friction of 0.5. If $g = 9.8 m / s^{2}$ , the smallest horizontal force that gives the weight an acceleration of $4 m / s^{2}$ is approximately

2.0 N

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2.5 N

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4.5N

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19.6N

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Solution

The correct option is C 4.5N

$F = μ m g + ma$

$4.9 = mg$

$m = \frac{4.9}{g} = \frac{4.9}{9.8} = 0.5 k g$

$∴ F = 0.5 \times 0.5 \times 9.8 + 0.5 \times 4$

$= 2.45 + 2 = 4.45 \approx 4.5 N$

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A weight of 4.9 N rests on a rough horizontal floor that has a co-efficient of friction of 0.5. If g=9.8m/s2, the smallest horizontal force that gives the weight an acceleration of 4m/s2 is approximately

The correct option is C 4.5N F = μmg + ma 4.9 = mg m=4.9g=4.99.8=0.5kg ∴F=0.5×0.5×9.8+0.5×4 = 2.45 + 2 = 4.45 ≈ 4.5 N

A weight of 4.9 N rests on a rough horizontal floor that has a co-efficient of friction of 0.5. If $g = 9.8 m / s^{2}$ , the smallest horizontal force that gives the weight an acceleration of $4 m / s^{2}$ is approximately

The correct option is C 4.5N

$F = μ m g + ma$

$4.9 = mg$

$m = \frac{4.9}{g} = \frac{4.9}{9.8} = 0.5 k g$

$∴ F = 0.5 \times 0.5 \times 9.8 + 0.5 \times 4$

$= 2.45 + 2 = 4.45 \approx 4.5 N$