Calculating Work Output, Input, and Effort

  • Thread starter Thread starter demode
  • Start date Start date
Click For Summary
To calculate work output for a barrel being rolled up a ramp, the resistance force (480 N) is multiplied by the height (2.0 m), resulting in 960 J. Work input is derived from the efficiency formula, yielding 24 J when efficiency is 40%. The effort applied is calculated using the work input equation, resulting in 1.5 N. For the car moving up a hill, work output is calculated as 240,000 J by multiplying the weight (16,000 N) by the height (15 m). The time taken is determined by dividing the distance (360 m) by the speed (12 m/s), leading to 30 seconds, and the engine's power output is calculated at 8,000 W.
demode
Messages
30
Reaction score
0
First Question

'A worker is rolling a 480-N barrel up a 16-m ramp onto a platform 2.0 m above the floor. The efficiency = arrangement is 40%. What is the work output, work input, and effort applied to the barrel..

To find work output, you would just multiply the resistance force (480) by the distance of resistance (2.0) and get 960 right?

To find work input , you would use the given information: Efficiency = Wout / Winput, so 40% = 960 / x = 24 J right?

TO find effort applied, you would just setup Winput = E * De, 24 = X * 16 = 1.5 N right?


Second Question

"16,000 N car is moving up a hill at 12 m/s, hill = 360 m long and 15 m high.. What work does the car's engine do in getting to the top, how long does it take, and what power does the engine do?"

I believe the first part is asking for Woutput, so Woutput = 16,000 * 15 = 240,000 J right?

To find out how long it takes, do we just divide the distance the car has to travel by the velocity? (360 / 12), or am i completely wrong there?
 
Physics news on Phys.org
To find the power of the engine, you would just divide the work output by the time it took to get to the top (240,000 / 30 = 8,000 W right?)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
17
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
1
Views
6K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
16K
  • · Replies 2 ·
Replies
2
Views
3K