Question

A body of weight 200 N is initially stationary on a 45 ° inclined plane. What distance along the inclined p lane must the body slide before it reaches a speed of 2m/sec. The

co-efficient of friction between the body and the plane is 0.1​

Answer 1

Answer:

41

IDK the correct answer, please check other resources too.

Answer 2

Answer:

The distance moved along the inclined plane is 0.31m.

Explanation:

Given the weight of a body,  $mg=200 N$

The angle of the inclined plane,  $\theta=45^o$

The final speed of the body, $v=2m/s$

The co-efficient of friction between the body and the plane, $\mu= 0.1$

The body is initially stationary, so initial velocity, $u=0$

The forces acting on the body are the frictional force(f) and the normal force(N) given by $f=mgsin\theta$ and $N=mgcos\theta$ respectively.

To make a body move, the balanced condition for force is

$f-\mu N=ma$  

$\implies mgsin\theta-\mu mgcos\theta=ma$

$\implies gsin\theta-\mu gcos\theta=a$

Substituting the given values, taking $g=10m/s^{2}$

$a=10\times sin45^{o} - 0.1\times 10\times cos45^{o}$

   $= \dfrac{10}{\sqrt{2} } - \dfrac{1}{\sqrt{2} }=\dfrac{9}{\sqrt{2} }=6.36m/s^{2}$

From the equation of motion for the distance travelled s, we have

$v^{2} -u^{2} =2as$

Substituting the values,

$2^{2} -0^{2} =2\times 6.36\times s$

$\implies s=\dfrac{4}{2\times 6.36} = 0.31m$

Therefore, the distance moved along the inclined plane is 0.31m.

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