A car initially at rest has a net force of 1000N (right) on it for 14 seconds. The brakes are then applied so that it skids to a stop. Determine the stopping distance of co-efficient = 0.35. The mass of the car is 800kg.
I have a rough idea about what to do, and I believe the answer is 70.24m.
Please/Thank You
Hmm...Well what I did was
acceleration = Fnet / mass
1000/800
= 1.25 m/s^2
Therefore since the car is accelerating for 14 seconds I got V = 17.5m/s (same as you).
Then we used forces to solve the rest.
Force Gravity = (800)(9.8) = 7840N
Force Friction = (7840)(0.35) = 2744N
Fnet = Fapplied - Ffriction
Fnet = 1000-2744
= -1744
Acceleration = -1744/800
= -2.18 m/s^2
V^2 = U^2 + 2ad
0^2 = 17.5^2 + 2(-2.18)d
= 70.24m?
Can you find any mistakes :S
I have a rough idea about what to do, and I believe the answer is 70.24m.
Please/Thank You
Hmm...Well what I did was
acceleration = Fnet / mass
1000/800
= 1.25 m/s^2
Therefore since the car is accelerating for 14 seconds I got V = 17.5m/s (same as you).
Then we used forces to solve the rest.
Force Gravity = (800)(9.8) = 7840N
Force Friction = (7840)(0.35) = 2744N
Fnet = Fapplied - Ffriction
Fnet = 1000-2744
= -1744
Acceleration = -1744/800
= -2.18 m/s^2
V^2 = U^2 + 2ad
0^2 = 17.5^2 + 2(-2.18)d
= 70.24m?
Can you find any mistakes :S