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Standard XII

Physics

Motion Under Gravity

A wooden box ...

Question

A wooden box of mass 8 Kg slides down an inclined plane of inclination $30^{o}$ to the horizontal with a constant acceleration of $0.4 m s^{- 2}$ . What is the force of friction between the box and inclined plane? $[g = 10 m s^{- 2}]$

56 . 8 N

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36 . 8 N

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65 . 6 N

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97 . 8 N

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Solution

The correct option is B 36 . 8 N
Given that,

Mass $m = 8 k g$

Inclination $θ = 30^{0}$

Acceleration $a = 0.4 m / s^{2}$

$g = 10 m / s^{2}$

We know that,

$m g sin θ - F_{f} = m a$

$m g sin 30^{0} - F_{f} = m a$

$8 \times 10 \times \frac{1}{2} - F_{f} = 8 \times 0.4$

$- F_{f} = 3.2 - 40$

$F_{f} = 36.8 N$

Hence, the frictional force is $36.8 N$

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A wooden box of mass 8 Kg slides down an inclined plane of inclination 30o to the horizontal with a constant acceleration of 0.4ms−2. What is the force of friction between the box and inclined plane?[g=10 ms−2]

The correct option is B 36 . 8 NGiven that, Mass m=8 kg Inclination θ=300 Acceleration a=0.4 m/s2 g=10 m/s2 We know that, mgsinθ−Ff=ma mgsin300−Ff=ma 8×10×12−Ff=8×0.4 −Ff=3.2−40 Ff=36.8N Hence, the frictional force is 36.8 N

A wooden box of mass 8 Kg slides down an inclined plane of inclination $30^{o}$ to the horizontal with a constant acceleration of $0.4 m s^{- 2}$ . What is the force of friction between the box and inclined plane? $[g = 10 m s^{- 2}]$