A block of mass 20 kg is hung with the help of idealstring, pulleys an

1.	A block of mass 20 kg is hung with the help of idealstring, pulleys and spring (Spring constantk = 1000 N/m) as shown in figure. If block is inequilibrium position then extension in the spring willbe (g = 10 ms-2)TO20 kg(1) 10 cmB 5 cm208 (2) 2.5 cm0 (4) 15 cm
Answer» Explanation: well where is the figure : / But I'm assuming the only force to be GRAVITATIONAL force F = mg = 20 X 10 = 200N k = 1000 N/m By Hooke's LAW, F = kx => 200 = 1000x => x = 0.2 m = 20 cm Hence the extension in the SPRING will be 20 cm

A block of mass 20 kg is hung with the help of idealstring, pulleys and spring (Spring constantk = 1000 N/m) as shown in figure. If block is inequilibrium position then extension in the spring willbe (g = 10 ms-2)TO20 kg(1) 10 cmB 5 cm208 (2) 2.5 cm0 (4) 15 cm

Answer»

Explanation:

well where is the figure : /

But I'm assuming the only force to be GRAVITATIONAL force

F = mg

= 20 X 10

= 200N

k = 1000 N/m

By Hooke's LAW,

F = k*x

=> 200 = 1000*x

=> x = 0.2 m

= 20 cm

Hence the extension in the SPRING will be 20 cm

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A block of mass 20 kg is hung with the help of idealstring, pulleys and spring (Spring constantk = 1000 N/m) as shown in figure. If block is inequilibrium position then extension in the spring willbe (g = 10 ms-2)TO20 kg(1) 10 cmB 5 cm208 (2) 2.5 cm0 (4) 15 cm​

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A block of mass 20 kg is hung with the help of idealstring, pulleys and spring (Spring constantk = 1000 N/m) as shown in figure. If block is inequilibrium position then extension in the spring willbe (g = 10 ms-2)TO20 kg(1) 10 cmB 5 cm208 (2) 2.5 cm0 (4) 15 cm