Work and Energy: Calculating Work in Various Situations

AI Thread Summary
The discussion focuses on calculating work in two scenarios involving a weight lifter and a car engine. For the weight lifter, the most work is done when lifting the barbell chest high (W1), followed by putting it down (W3), and no work is done while holding it (W2) due to lack of displacement. In the second scenario, the work done by the engine on a 1200 kg car accelerating at 1.2 m/s² over 150 meters is calculated using the formula Wd = F × d, resulting in 216,000 J. The calculations and application of the work equations are confirmed to be correct. Understanding the concepts of force, displacement, and the equations for work is essential for solving these types of problems.
Weatherkid11
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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
 
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I think you are right for both questions, you've used the equations:

Wd=F \times d and
F=m\times a

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 \frac{m}{s^2} (meters per second squared)

I think you have interpreted and used these equations correctly.
 
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