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Please let me know if I am on the right track for these problems. There are two other ones i have added. PLease let me know if i am right for all four. Thanks a bunch!

1) A weight lifter picks up a barbell and

1. lifts it chest high

2. holds it for 5 minutes

3. puts it down.

Rank the amounts of work W the weight lifter performs during this three operations. Label the quantities as W1, W2, and W3. Justify you ranking order. ---> I think that the most amount of work is when it is lifted chest high, since there is a force and displacement involved. After that, would me when he puts it down, becuase gravity helps, and there is a displacement. When he holds it up for five min there is no work becuase there is no displacement.

2)Estimate the amount of work the engine performed on a 1200 kg car as it accelerated at 1.2 m/s2 over a 150 meter distance. ----> Well since the equation for work is force times distance, it would be (1.2 m/s2)(1200kg)(150m) So the total work would be 216,000 J. But this seems to easy, so i think i am wrong

3) Image for this promblem: http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter7/medialib/chap7gifs/chap7_wu_4.gif [Broken]

A real world roller-coaster released at point A and coasting without external power would traverse a track somewhat like the figure above. Friction is not negligible in the real world. If we call the potential energy at point A relative to the ground Uo what can you say about the potential and kinetic energies UB, UB at point B and the potential and kinetic energies UC, UC at point C.----> Here is what I think: KA + UA = ( KB + UB ) + ( KC + UC ) + Ffr

So plugging in the variables, we get: mg(ha) = ( 1/2mvb^2 + mg(hb)) + ( 1/2mvb^2 + mg(hc))+Ffr

where:

m=mass

vb= velocity at point B

g= gravity

hb= height of point B

hc= height of point C

Ffr= Frictional force

4) A gallon of gasoline contains about 1.3 x 10^8 joules of energy. A 2000 kg car traveling at 20 m/s skids to a stop. Estimate how much gasoline it will take to bring the car back to the original speed? To complicate matters further, consider the fact that only about 15% of the energy extracted from gasoline actually propels the car. The rest gets exhausted as heat and unburnt fuel. ---> Wow... this one i am totally lost. Please help

1) A weight lifter picks up a barbell and

1. lifts it chest high

2. holds it for 5 minutes

3. puts it down.

Rank the amounts of work W the weight lifter performs during this three operations. Label the quantities as W1, W2, and W3. Justify you ranking order. ---> I think that the most amount of work is when it is lifted chest high, since there is a force and displacement involved. After that, would me when he puts it down, becuase gravity helps, and there is a displacement. When he holds it up for five min there is no work becuase there is no displacement.

2)Estimate the amount of work the engine performed on a 1200 kg car as it accelerated at 1.2 m/s2 over a 150 meter distance. ----> Well since the equation for work is force times distance, it would be (1.2 m/s2)(1200kg)(150m) So the total work would be 216,000 J. But this seems to easy, so i think i am wrong

3) Image for this promblem: http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter7/medialib/chap7gifs/chap7_wu_4.gif [Broken]

A real world roller-coaster released at point A and coasting without external power would traverse a track somewhat like the figure above. Friction is not negligible in the real world. If we call the potential energy at point A relative to the ground Uo what can you say about the potential and kinetic energies UB, UB at point B and the potential and kinetic energies UC, UC at point C.----> Here is what I think: KA + UA = ( KB + UB ) + ( KC + UC ) + Ffr

So plugging in the variables, we get: mg(ha) = ( 1/2mvb^2 + mg(hb)) + ( 1/2mvb^2 + mg(hc))+Ffr

where:

m=mass

vb= velocity at point B

g= gravity

hb= height of point B

hc= height of point C

Ffr= Frictional force

4) A gallon of gasoline contains about 1.3 x 10^8 joules of energy. A 2000 kg car traveling at 20 m/s skids to a stop. Estimate how much gasoline it will take to bring the car back to the original speed? To complicate matters further, consider the fact that only about 15% of the energy extracted from gasoline actually propels the car. The rest gets exhausted as heat and unburnt fuel. ---> Wow... this one i am totally lost. Please help

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