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a block of mass 200g is suspended through a vertical spring.the spring is stretched by 1 cm when the block was in equilibrium.a particle of mass 120g is dropped on the block from the hieght of 45cm .the particle sticks to the block after the impact.find the maximum extension in the string.

a block of mass 200g is suspended through a vertical spring.the spring is stretched by 1 cm when the block was in equilibrium.a particle of mass 120g is dropped on the block from the hieght of 45cm .the particle sticks to the block after the impact.find the maximum extension in the string.

Grade:11

2 Answers

Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
9 years ago
Hi,

In this question, apply law of conservation of energy.
Decrease in potential energy of blocks = Increase in PE of spring.

It would be quadratic equation in x , solve it to get the maximum extension.

Thanks.
Smw
11 Points
6 years ago
Since when the block is in equlibrium its all the forces are balancedlet spring constant be k and extension be x0kx0 = mg k = mg/x0 = 200 x 10-3 x 10/10-2 = 200 N/m .......................................1>now when the mass is dropped from some height now applying work energy theorm work done by all the forces equal to its change in kinectic energy of systeminitially system has k.E =0 and after maximum extension the system again comes to rest K.E finally = 0∆k.E = Wgravity + W​spring force0 = mgh +( M+m)gx - 1/2 kx2 (here x is max extension m , M are mass of 120g and 200 gram, h = 45cm,k =200 N/m)0 = 120x 10-3x 10 x 45 x 10-2 + (​ 200+120) x10-3 x 10x -1/2 x 200 x20 = 54 x 10-2 + 3.2x - 100x20= 0.54 +3.2x -100x20 = 100x2 - 3.2x -0.54​x =−(−3.2)+3.22 − 4×100×−0.54√2×1003.2+10.24+216√2003.2+15.04200 = 18.24200 =0.0912 m = 9.1 cm

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