NEET and AIPMT NEET Physics Work,Energy and Power MCQ Questions Solved


A spring 40 mm long is stretched by the application of a force. If 10 N force required to stretch the spring through 1 mm, then work done in stretching the spring through 40 mm is

(a) 84J

(b) 68J

(c) 23J

(d) 8J

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Since force is same

F= k1x1 = k2x2 So, x1x2= k2k1Also, U1U2=12k1x2112k2x22 =Fk2Fk1=k2k1

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(1)

When the string is fully compressed.
K.E. of the block = P.E. of the spring.

12mv2 = 12 Kx212×25×32 = 12×100×x2x=1.5 m

When the block comes to the original psotion, it will gain its K.E. from spring and the speed of the block will be same as before.
 V=-3 m/s

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(2)

          

While crossing the equilibrium positionT=mg+mv2r=mg+ml2gl1-cosθ   =mg+2mg-2mg cosθ=mg3-2cosθ     3-2 cosθ=29401509.8=2     2 cosθ=1  cosθ=12θ=60°

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