InterviewSolution
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InterviewSolution
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In figue `k=100Nm^-1 M=1kg and F=10N`. a. Find the compression of the spring in the equilibrium position. b. A sharp blow by some xternal agent imparts a speed of `2ms^-1` to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at ths instant. c. find the time period of the resulting simple harmonic motion. d. Find the ampitude. e. Write the potential energy of the spring when the block is at the left extreme. f. Write teh potential energy of the spring when teh block is at the right extreme. The anwer of b, e, and f are different. Explain why this does not violate the principle of conservation of energy. |
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Answer» Correct Answer - A::B::C::D Given `k=100BN/m, m=1kg and F=10 N a. In the equilibrium position `Compression `=delta=F/k=10/100` `=0.1m=10cm` b. The below imparts a speed of 2 m/s to the block towards left. `:.+P.E.+K.E.=1/2kdelta^2+1/2Mv^2` `=1/2x100xx(0.1)^2+1/2x1xx4` `=0.5+2=2.5J` `c. Time period `2pisqrt(M/k)` `=2pi sqrt(M/100)=pi/5 sec` d. Let the amplitude be x whigh means the distasnce betwen the mean position and the extreme position. So in the extreme position, compression of the spring is `(x-K)` Since in SHM the totl energy remains constant, `1/2k(x+delta)^2=1/2kdelta^2+1/2mv^2+Fx` `=2.5+10x [because 1/2kdelta^2+1/2mv^2=2.5]` so, `50(x+0.1)^2=2.5+10x` `:. 50x^2+0.5+10x=2.5+10x` `=:. 50x^2=2` `rarr x^2=2/50=4/100` `rarr x=2/50x=20m` e. potential energy at the left extreme is given by `P.E. =1/2k(x+delta)^2` `=1/2x100xx(0.1+0.2)^2` `=50xx(0.09)=4.5J` f. Potential energy at the right extreme is given by `P.E. 1/2k(x+delta)^2-F(2x)` `[2x`=distance between two extreme)` `=4.5-10(0.4)=0.5J` The different values is b, e, and f do not violate law of conservastion of energy as the work is done by the exterN/Al force 10 N. |
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