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Question

A 0.5 kg block slides from the point A on a horizontal track with an initial speed 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/m. The part AB of the Track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. $$\displaystyle \left ( g= 10 m/s^{2} \right )$$


Solution

As the track AB is frictionless, the block moves this distance without loss in its initial $$\displaystyle KE= \frac{1}{2}mv^{2}= \frac{1}{2}\times 0.5\times 3^{2}= 2.25 J.$$ In the path BD as friction is present, so work done against friction
$$\displaystyle = \mu _{k}mgs= 0.2\times 0.5\times 10\times 2.14= 2.14J$$
So, at D the KE of the block is=2.25-2.14=0.11J
Now, if the spring is compressed by x
$$\displaystyle 0.11= \frac{1}{2}\times k\times x^{2}+\mu _{k}mgx$$
i.e., $$\displaystyle 0.11= \frac{1}{2}\times 2\times x^{2}+0.2\times 0.5\times 10x$$
or $$\displaystyle x^{2}x-0.11= 0$$
which on solving gives positive value of x=0.1 m
After moving the distance x=0.1 m the block comes to rest. Now the compressed spring exerts a force:
$$\displaystyle F= kx= 2\times 0.1= 0.2N$$
On the block while limiting frictional force between block and track is $$\displaystyle f_{L}= \mu _{s}mg$$
$$\displaystyle = 0.22\times 0.5\times 10= 1.1N.\:Since,\:F< f_{L}.$$ The block will not move back. So, the total distance moved by the block
$$\displaystyle = AB+BD+0.1$$
$$=2+2.14+0.1$$
$$=4.24m$$

Physics

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