A block A of mass 2 kg starts slipping down the inclined plane of incl – InterviewSolution

1.

A block A of mass 2 kg starts slipping down the inclined plane of inclination37° from a point 4.8 m away from the spring of constant k=1000 N/m . The coefficient of kinetic friction between the block and the inclined plane is 0.5. If the maximum compression in the spring is x_max , then find the value of 10x_max , in metre

Answer»

Topic :-

Mechanics

Given :-

A block A of MASS 2 kg starts slipping down the inclined plane of inclination 37° from a point 4.8 m away from the spring of CONSTANT k = 1000 N/m. The coefficient of kinetic friction between the block and the inclined plane is 0.5.

Formula to be Used :-

$<klux>WORK</klux>\:done\:by\:spring=\dfrac{1}{2}kx^2$

x is maximum compression in spring.

Frictional force when a block slips down =

$\mu mgcos\theta$

Force with which block slips down =

$mgsin\theta$

$Work\:Done=Force \times Displacement$

Solution :-

It is given that,

Mass of block, m = 2 kg

Inclination of plane = 37°

DISTANCE traversed by block = Distance between block and spring + Maximum Compression in spring

Distance traversed by block = ( 4.8 + x ) metres

Spring Constant, k = 1000 N/m

Kinetic friction between block and plane = 0.5

Now, calculate work done by block,

Work done = Force × Displacement

$Work\:done = ( mgsin\theta - \mu mgcos\theta) \times Displacement$

$Work\:done =\left ( (2)(10)sin37^{\circ} -\dfrac{1}{2}(2)(10)cos37^{\circ} \right ) \times (4.8+x)$

$Work\:done=\left (20 \times \dfrac{3}{5} - \dfrac{1}{2} \times 20 \times \dfrac{4}{5} \right )\times (4.8+x) \: Joules$

$Work\:done=\left (12 - 8 \right )\times (4.8+x) \: Joules$

$Work\:done=4 \times (4.8+x) \: Joules$

$Work\:done=19.2+4x\:Joules$

Work done by spring,

$Work\:done\:by\:spring=\dfrac{1}{2}kx^2$

As we know, for maximum compression whole energy of block will shift into spring energy. So, we will equate both equations.

So,

$\dfrac{1}{2}kx^2=19.2+4x$

$\dfrac{1}{2} \times 1000 \times x^2=19.2+4x$

$500x^2=19.2+4x$

$500x^2-4x-19.2=0$

Apply, quadratic formula,

$x=\dfrac{4 \pm \sqrt{16-4(500)(-19.2)}}{1000}$

$x=\dfrac{4 \pm \sqrt{38416}}{1000}$

$x=\dfrac{4 \pm 196}{1000}$

$x=0.2\:\:\:and\:-0.192$

x can't be NEGATIVE so we reject negative value of 'x'.

So,

$10x_{max}=10 \times 0.2\:metres$

$10x_{max}=2\:metres$

Answer :-

So, ten times of maximum compression in spring is 2 metres.

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