Asked by Jenna
Given: g=9.81 m/s square.
The parachute on a race car that weighs 8042 N opens at the end of a quarter-mile run when the car is traveling 35 m/s. What net retarding force must be supplied by the parachute to stop the car in a distance of 1020 m? Answer in units of N.
I have no idea what formula I should use or how to solve this problem.
The parachute on a race car that weighs 8042 N opens at the end of a quarter-mile run when the car is traveling 35 m/s. What net retarding force must be supplied by the parachute to stop the car in a distance of 1020 m? Answer in units of N.
I have no idea what formula I should use or how to solve this problem.
Answers
Answered by
bobpursley
retarding force*distance=Kineticenergyofcar
Answered by
Jenna
so it would be:
(retarding force)(1020m)= (1/2)(8042N)(35 m/s)
and the retarding force= 137.98?
(retarding force)(1020m)= (1/2)(8042N)(35 m/s)
and the retarding force= 137.98?
Answered by
BLT
You would first find the mass of the car using Weight = m*g
Once you have found the mass, you can solve for "a" in finding Fnet (Fnet= m*a) so (Fnet= 821*a)
Because Fnet= m a you can rearrange the equation and substitute this into the following equation:
(a)
V^2= Vi^2 + 2(Fnet/m)(X)
0 = (35^2) + 2(Fnet/821) (1020)
Therefore the Net Force will be approximately 493 N in the opposing direction. (Because it is an opposing force, the answer will come out negative)
Once you have found the mass, you can solve for "a" in finding Fnet (Fnet= m*a) so (Fnet= 821*a)
Because Fnet= m a you can rearrange the equation and substitute this into the following equation:
(a)
V^2= Vi^2 + 2(Fnet/m)(X)
0 = (35^2) + 2(Fnet/821) (1020)
Therefore the Net Force will be approximately 493 N in the opposing direction. (Because it is an opposing force, the answer will come out negative)
Answered by
DennisSeell
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