I am having some trouble with this problem from my homework. It isfrom Physics (

Question

I am having some trouble with this problem from my homework. It isfrom Physics (8th), Cutnell, Chapter 10 Problem 39. It reads asfollows:

-A block rests on a frictionless horizontal surface and is attachedto a spring. When set into simple harmonic motion, the blockoscillates back and forth with an angular frequency of 7.5 rad/s.The drawing shows the position of the block when the spring isunstrained. This position is labeled ''x = 0 m.'' Thedrawing also shows a small bottle located 0.080 m to the right ofthis position. The block is pulled to the right, stretching thespring by 0.050 m, and is then thrown to the left. In order for theblock to knock over the bottle,it must be thrown with a speedexceeding v0. Ignoring the width of the block,find v0.

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Explanation / Answer

Given : The block oscillates back and forth with an angular frequencyof () = 7.5 rad/s. The drawing also shows a small bottle located (xf) = 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by(xo) = 0.050 m, and is then thrown to theleft Therefore now we have to find (vo)=?                       From the conservation of mechanical energy
It states that the final total mechanical energyEf is equal to the initial totalmechanical energy E0. The expression for the total mechanical energy for aspring/mass system is given by                       (1/2)mvf2+(1/2)If2+mghf+(1/2)kxf2 =(1/2)mvo2+(1/2)Io2+mgh0+(1/2)kxo2 Here the block is not rotating than the angualr speedf and 0 are zero And moreover, the block reaches the bottle with a final speedof vf = 0 m/s when the block islaunched with the minimum initial speedv0. In addition, the surface is horizontal, so that the final andinitial heights, hf andh0, are the same. Thus, the above expression becomes as                                         (1/2)kxf2 = (1/2)mvo2 +(1/2)kxo2                                                 kxf2 = mvo2 +kxo2                                                 mvo2= k(xf2-xo2)                                                   vo2 =(k/m)(xf2-xo2)..................(1) As we know that the angular frequency () =k/m                                                       2 = k/m ................(2) Susbtituting the equation (2) in equation (1) weget                                                       vo2 =2(xf2-xo2) Now suqaring on both the sides we get vo as                                                       vo =(xf2-xo2)                                                            =(7.5rad/s)[ (0.080 m)2 -(0.050m)2                                                                  =(7.5rad/s)[ (0.0064)-(0.0025)                                                                         =(7.5rad/s) ( 0.0039)                                                           = (7.5rad/s)(0.062449)                                                           = 0.468m/s I hope it helpsyou The block oscillates back and forth with an angular frequencyof () = 7.5 rad/s. The drawing also shows a small bottle located (xf) = 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by(xo) = 0.050 m, and is then thrown to theleft Therefore now we have to find (vo)=?                       From the conservation of mechanical energy
It states that the final total mechanical energyEf is equal to the initial totalmechanical energy E0. The expression for the total mechanical energy for aspring/mass system is given by                       (1/2)mvf2+(1/2)If2+mghf+(1/2)kxf2 =(1/2)mvo2+(1/2)Io2+mgh0+(1/2)kxo2 Here the block is not rotating than the angualr speedf and 0 are zero And moreover, the block reaches the bottle with a final speedof vf = 0 m/s when the block islaunched with the minimum initial speedv0. In addition, the surface is horizontal, so that the final andinitial heights, hf andh0, are the same. Thus, the above expression becomes as                                         (1/2)kxf2 = (1/2)mvo2 +(1/2)kxo2                                                 kxf2 = mvo2 +kxo2                                                 mvo2= k(xf2-xo2)                                                   vo2 =(k/m)(xf2-xo2)..................(1) As we know that the angular frequency () =k/m                                                       2 = k/m ................(2) Susbtituting the equation (2) in equation (1) weget                                                       vo2 =2(xf2-xo2) Now suqaring on both the sides we get vo as                                                       vo =(xf2-xo2)                                                            =(7.5rad/s)[ (0.080 m)2 -(0.050m)2                                                                  =(7.5rad/s)[ (0.0064)-(0.0025)                                                                         =(7.5rad/s) ( 0.0039)                                                           = (7.5rad/s)(0.062449)                                                           = 0.468m/s I hope it helpsyou                                                           = (7.5rad/s)(0.062449)                                                           = 0.468m/s I hope it helpsyou I hope it helpsyou

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