Work done against resistive forces by comparing Ek and work done

AI Thread Summary
The discussion centers on calculating the work done against resistive forces for a car accelerating from 0 to 26.8 m/s with a constant engine thrust of 3500 N. The kinetic energy gained by the car is compared to the total work done by the engine, leading to a calculated work done against resistive forces of 110,000 J. There is a correction regarding the calculation of work done, emphasizing the importance of accurate unit representation. The final answer confirms the work done against resistive forces as 110,000 J.
g9WfI
Messages
13
Reaction score
4
Homework Statement
A car of mass 900kg accelerates from 0 to 26.8 m/s in 9.2 s. The thrust from the engine is constant at 3500 N during this period. By comparing the kinetic energy gained by the car with the total work done by the engine, calculate the work done against resistive forces.
Relevant Equations
suvat, kinetic energy, work done
IMG_86F1E608F5D0-1.jpeg

Missed units in the photo - J

Answer is 110 000 J
 
Physics news on Phys.org
g9WfI said:
Homework Statement:: A car of mass 900kg accelerates from 0 to 26.8 m/s in 9.2 s. The thrust from the engine is constant at 3500 N during this period. By comparing the kinetic energy gained by the car with the total work done by the engine, calculate the work done against resistive forces.
Relevant Equations:: suvat, kinetic energy, work done

View attachment 284260
Missed units in the photo - J

Answer is 110 000 J
3500 * 123.48 = 431480, not 43300
 
Steve4Physics said:
3500 * 123.48 = 431480, not 43300
Thank you so much!
 
Reply
  • Like
Likes Steve4Physics

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Work done by a balloon that is rising
Replies
8
Views
725
Internal Energy of Gas and Work done
Replies
10
Views
2K
Regarding the work done against air resistance
Replies
2
Views
3K
Why is the work done double its expected value? (conveyer belt)
2
Replies
58
Views
5K
Forces involved in principle of virtual work
Replies
2
Views
709
Calculating work done by a gas
Replies
2
Views
716
How can "work" done by a force be negative?
Replies
19
Views
1K
Work done by a force down a ramp
Replies
11
Views
2K
How do I find the work done for a 100 N downward force?
Replies
4
Views
2K
My Mistake? Understanding Friction Force & Work Done on Snowy & Icy Surfaces
Replies
29
Views
2K

Hot Threads

Recent Insights

Back
Top