What is the Correct Force Needed to Slow a Drag Racer with a Chute?

[physics114]

Correct?

1. Given Problem
A drag racer crosses the finish line doing 203mi/h and promptly deploys her drag chute (the small parachute used for braking).
(a)What force must the drag chute exert on the 878kg car to slow it to 45.2 mi/h in a distance of 199m?
(b)Describe strategy.


2. Formulas
F=ma
vf^2=vi^2+2aΔx
1mi=1609m
1hr=3600s


3. Attempted Solution
b) Strategy
First convert mi/hr units into m/s. Find acceleration by using vf^2=vi^2+2aΔx. Then use the acceleration found and finding Force using Newton's second law.

a) Calculations
Converted
203mi/hr = 90.279m/s
45.2mi/hr = 20.2019m/s

Plug into Motion Equation for Acceleration
(vf^2-vi^2)/(Δx) = 2a
[(90.279)^2 - (20.2019)^2] / [(199m)] = 2a
(408.117-8231.75) / (199m) = 2a
-39.3147 = 2a
a = -39.3127/2
a = -19.6574 m/s^2

Use Newton's Law
F=ma
F= (878kg)(-19.6574m/s^2)
F= -17259.2N

 

[NeedsHelp1212]

90.279 squared is 8150.3 not 408.117, really not sure what you did there. It should be 8150.3 minus 408=7742, divide that by 199=38.9, divide that by 2= + 19.5 m/s squared

Net force should be 19.5 times the mass (878)= 17,079 N

 

[physics114]

Thank you =] Made the correction but apparently my answer has the wrong sign? Answer should be in x direction which I thought was what I was finding. Slightly confused =/

 

[NeedsHelp1212]

Um not sure, Physics confuses me...lol

 

[rwisz]

Yes, this is the correct solution. To slow the car from 203mi/h to 45.2mi/h in a distance of 199m, the drag chute must exert a force of -17259.2N on the car. The strategy used is correct, as it involves converting units, using motion equations to find acceleration, and then using Newton's second law to calculate the force. Great job!