Block of mass 2 kg is dropped vertically from a height of 40 cm onto a spring. the force constant of the spring is 1960 nm-1. calculate the maximum compression of the
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hii mate!!
.
When the mass is slowly lowered, the spring will extend till the gravitational force on the mass is balanced by the restoring force of the spring. Such that, mg = kd => d = mg/k
When the mass is allowed to fall freely it will attain a kinetic energy which will put the mass in an oscillatory about the mean extension of the spring which is also ‘d’.
2.
Initial PE of the block is = mgh = (2)(10)(0.4) = 8 J [g = 10 m/s2]
When it hits the spring the compression of the spring is such that the mechanical energy of the block is equal to the potential energy stored in the spring.
hope it helps !!☺☺☺
So, ½ kx2 = 8
=> x2 = (2)(8)/(1960) = 0.008
=> x = 0.09 m = 9 cm
.
When the mass is slowly lowered, the spring will extend till the gravitational force on the mass is balanced by the restoring force of the spring. Such that, mg = kd => d = mg/k
When the mass is allowed to fall freely it will attain a kinetic energy which will put the mass in an oscillatory about the mean extension of the spring which is also ‘d’.
2.
Initial PE of the block is = mgh = (2)(10)(0.4) = 8 J [g = 10 m/s2]
When it hits the spring the compression of the spring is such that the mechanical energy of the block is equal to the potential energy stored in the spring.
hope it helps !!☺☺☺
So, ½ kx2 = 8
=> x2 = (2)(8)/(1960) = 0.008
=> x = 0.09 m = 9 cm
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