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  • A block of weight 100 N is pushed by a force F on a horizontal rough plane and moves with an acceleration

A block of weight 100 N is pushed by a force F on a horizontal rough plane and moves with an acceleration 1 \mathrm{~m} / \mathrm{s}^2, when force is doubled its acceleration becomes 10 \mathrm{~m} / \mathrm{s}^2. The coefficient of friction is \left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)

Option: 1

0.2

 


Option: 2

0.4


Option: 3

0.6

 


Option: 4

 0.8


Answers (1)

best_answer

 Weight, W =100 N
Acceleration, a=1 \mathrm{~m} / \mathrm{s}^2
Acceleration when force is double, a=10 \mathrm{~m} / \mathrm{s}^2

Now, Let the External Force be 'F' and the friction force be ' \mathrm{f_s}'.
Then,
\begin{aligned} & \mathrm{F}-\mathrm{f}_{\mathrm{s}}=\mathrm{ma} \\ & \mathrm{F}-\mu \mathrm{N}=\mathrm{ma} \\ & \mathrm{F}-\mu 100=10 \times 1 \\ & \mathrm{~F}-\mu 100=10 \ldots .(\mathrm{I}) \end{aligned}

We know, When Force is doubled, Then,
\begin{aligned} & 2 \mathrm{~F}-\mathrm{f}_{\mathrm{s}}=\mathrm{ma} \\ & 2 \mathrm{~F}-\mu \mathrm{N}=\mathrm{ma} \\ & 2 \mathrm{~F}-\mu 100=10 \times 10 \\ & 2 \mathrm{~F}-\mu 100=100 \ldots \text { (II) } \end{aligned}

Now, from equations (I) and (II)
\mathrm{F}=90 \mathrm{~N}
Now, put the value of F in equation (I)
\begin{aligned} & 90-\mu 100=10 \\ & -\mu 100=10-90 \\ & \mu=\frac{80}{100} \\ & \mu=0.8 \end{aligned}
Hence, the coefficient of friction is 0.8

Posted by

Divya Prakash Singh

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