Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The total mass is 2300kg (weight = 5072lbs). Find the size of the force due to wind and rolling friction when the van speed is 44.86mph (1mph= 0.447m/s).

image

I am really at a loss of what to do here.

Part 2:

How much net power, in hp (1 hp=746W) must the engine deliver to maintain a speed of 44.86mph (20.05m/s) (Neglect losses involved in delivering the power).

share|improve this question

closed as off-topic by Brandon Enright, John Rennie, Kyle Kanos, Chris White, jinawee Mar 13 at 15:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Brandon Enright, John Rennie, Kyle Kanos, Chris White, jinawee
If this question can be reworded to fit the rules in the help center, please edit the question.

    
Are you meant to calculate the acceleration from the graph? –  Jedediyah Mar 12 at 5:07
    
no... Bad tag, I fixed it. –  user2840324 Mar 12 at 5:09
add comment

1 Answer 1

up vote 0 down vote accepted

Based on the graphed data, the acceleration appears constant and I estimate it at $$ a = \frac{40-55}{25-5} = -0.75 \text{ mph/s} = -0.33528 \text{ m / (s$^2$)}$$ You know the mass is 2300 kg, so you can just plug in to get the force of friction using $$ F = ma $$

As for the power, I believe (please check this) that in your 1D case, the power is given as $P = Fv$ where $v$ is the velocity.

share|improve this answer
    
Thank you very much. –  user2840324 Mar 12 at 6:21
    
Please don't post complete answers to homework-like questions in the future, as mentioned in our homework policy. –  David Z Mar 14 at 18:19
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.