NEET Physics Work,Energy and Power 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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(3)

P= F.vP=ma.vP=mvdvdtPdt=mvdvPdt=mvdvP.t=mv22v=2Ptmdx=2Ptmdtdx=2Pmtdtx=2Pm×23t3/2

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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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(1)

                                          

At the point, where the block leaves the circle T=0

Using conservation of energy from A to B-

Ui+Ki = Uf+KfO+12mu2 = mgl+l sinθ+12mv2   v2=u2-2gl+lsinθ                ............iAt point B-   T=0 mg sinθ = mv2l      v2=gl sinθput it in i   lg sinθ = xgl-2gl-2gl sinθ   3l sinθ=x-2l       sinθ=x-2/3h=l+lsinθ=l+lx-23h=lx+13

v=xglh=x+13lwhere h is height where body leave the circle.h=72+13l=3l2

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Tmax=mv12R+mg = mv12+mgRRTmin=mv22R-mg = mv22-mgRRAs,    TmaxTmin = 4=v12+gRv22-gR             4v22-gR = v12+gR                     v12 = 4v22-5gR                ............iConserving energy from L to H -          12mv12 = 12mv22 + 2mgR               v12 = v22 + 4gR                        ...........iiOn comparing both equations-            4v22-5gR = v22 + 4gR            3v22 = 9gR               v22 = 3×10×103            v2=10 m/s

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