In spring constant of the spring is 100 `N//m` Extension produced in the spring is:
`(g=10 m//s^(2))`

A mass M is suspended as shown in fig. The system is in equilibrium. Assume pulleys to be massless. K is the force constant of the spring. The extension produced in the spring is given by

A mass M is suspended as shown in fig. The system is in equilibrium. Assume pulleys to be massless. K is the force constant of the spring. The extension produced in the spring is given by

Natural length of a spring is 60 cm and its spring constant is 4000 N/m. A mass of 20 kg is hung from it. The extension produced in the spring is (Take, g = 9.8 m//s^(2) )

Natural length of a spring is 60 cm, and its spring constant is 4000 N/m, A mass of 20 kg is hung from it. The extension produced in the spring is (take g=9.8m//s^(2) )

Natural length of a spring is 60 cm and its spring constant is 4000 N/m. A mass of 20 kg is hung from it. The extension produced in the spring is (Take, g = 9.8 m//s^(2) )

Figure shows a force compression curve of a spring. A body of mass 5kg moving with the velocity of 8m//s hits the spring. Calculate the force constant of the spring and also the compression produced in the spring when the body hits it.

The system is released from rest with both the springs in unstretched positions. Mass of each block is 1 kg and force constant of each spring is 10 N//m . Extension of horizontal spring in equilibrium is

The system is released from rest with both the springs in unstretched positions. Mass of each block is 1 kg and force constant of each spring is 10 N//m . Extension of horizontal spring in equilibrium is

Force constant of a spring is 100 N//m . If a 10kg block attached with the spring is at rest, then find extension in the spring. ( g= 10 m//s^(2) )