Work and Energy: Calculating Work in Various Situations

[Weatherkid11]

Please let me know if I am on the right track for these problems.
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

 

[sanitykey]

I think you are right for both questions, you've used the equations:

[tex]Wd=F \times d[/tex] and
[tex]F=m\times a[/tex]

Where:

Wd is work done in Nm (Newton meters)
F is the force in N (Newtons)
d is the distance in m (meters)
m is the mass in kg (kilogrammes)

a is the acceleration in [tex]\frac{m}{s^2}[/tex] (meters per second squared)

I think you have interpreted and used these equations correctly.

 

[quicknote]

.

Your reasoning for the first problem is correct. The weight lifter does the most work when lifting the barbell to chest height, as there is both a force (from the weight of the barbell) and a displacement involved. The least amount of work is done when holding the barbell for 5 minutes, as there is no displacement and therefore no work being done.

For the second problem, your calculation is correct. The work done by the engine is equal to the force (mass x acceleration) times the distance. In this case, the force is the weight of the car (mg) and the distance is 150 meters. So the total work done is 216,000 J. Good job!