A 10 kg mass is resting on a horizontal surface and horizontal force of 80 N is applied- If - -0-2- the ratio of acceleration without friction and with friction is g - 10ms-2

The correct option is B 43 a_1/a_2= #160;F/m/F μ mg/m a_1 = F/m = 80/10 = 8 m/s^2 a_2 = F μ #160; mg/m = 80 0.2× 10× 10/10 = 6 m/s^2 ...

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Standard XII

Physics

Coefficients of Friction

A 10 kg mas...

Question

A $10$ kg mass is resting on a horizontal surface and horizontal force of $80$ N is applied. If $μ = 0.2$ , the ratio of acceleration without friction and with friction is ( $g = 10 m s^{- 2}$ )

\frac{3}{4}

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\frac{4}{3}

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\frac{1}{2}

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2

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Solution

The correct option is B $\frac{4}{3}$
$\frac{a_{1}}{a_{2}} = \frac{\frac{F}{m}}{\frac{F - μ m g}{m}}$
$a_{1} = \frac{F}{m} = \frac{80}{10} = 8 m / s^{2}$
$a_{2} = \frac{F - μ m g}{m} = \frac{80 - 0.2 \times 10 \times 10}{10} = 6 m / s^{2}$
$∴ \frac{a_{1}}{a_{2}} = \frac{8}{6} = \frac{4}{3}$

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A 10 kg mass is resting on a horizontal surface and horizontal force of 80 N is applied. If μ=0.2, the ratio of acceleration without friction and with friction is (g=10ms−2)

The correct option is B 43a1a2=FmF−μmgma1=Fm=8010=8 m/s2a2=F−μmgm=80−0.2×10×1010=6 m/s2∴a1a2=86=43

A $10$ kg mass is resting on a horizontal surface and horizontal force of $80$ N is applied. If $μ = 0.2$ , the ratio of acceleration without friction and with friction is ( $g = 10 m s^{- 2}$ )

The correct option is B $\frac{4}{3}$
$\frac{a_{1}}{a_{2}} = \frac{\frac{F}{m}}{\frac{F - μ m g}{m}}$
$a_{1} = \frac{F}{m} = \frac{80}{10} = 8 m / s^{2}$
$a_{2} = \frac{F - μ m g}{m} = \frac{80 - 0.2 \times 10 \times 10}{10} = 6 m / s^{2}$
$∴ \frac{a_{1}}{a_{2}} = \frac{8}{6} = \frac{4}{3}$