How Strong is the Stopping Force for a 1.5×10^4 N Car?

  • Thread starter Thread starter bearhug
  • Start date Start date
  • Tags Tags
    Force Magnitude
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 6K views
bearhug
Messages
78
Reaction score
0
A car that weighs 1.5×104 N is initially moving at a speed of 45 km/h when the brakes are applied and the car is brought to a stop in 12 m. Assuming that the force that stops the car is constant, find the magnitude of that force.

I found the mass to be 1.53e3 kg then I figured now I need to find the acceleration using the information that I have so I used the equation
Vf^2 = Vi^2 + 2a(Xf-Xi)
Vi = 12.5 m/s
Vf = 0 m/s
Xf = 12m
Xi= 0 m

Then plug the acceleration value into the equation F= ma but apparently that's wrong. Is this the correct method?
 
Physics news on Phys.org
bearhug said:
A car that weighs 1.5×104 N is initially moving at a speed of 45 km/h when the brakes are applied and the car is brought to a stop in 12 m. Assuming that the force that stops the car is constant, find the magnitude of that force.

I found the mass to be 1.53e3 kg then I figured now I need to find the acceleration using the information that I have so I used the equation
Vf^2 = Vi^2 + 2a(Xf-Xi)
Vi = 12.5 m/s
Vf = 0 m/s
Xf = 12m
Xi= 0 m

Then plug the acceleration value into the equation F= ma but apparently that's wrong. Is this the correct method?
Method is ok. [itex]a = \Delta v/\Delta t \text{ where } \Delta t = 2d/v_i[/itex] Since final velocity is 0, you have: a = v^2/2d

You are in effect using energy: [itex]KE = F*d = \frac{1}{2}mv^2[/tex] which is a little more direct:<br /> <br /> [tex]F = mv^2/2d[/tex]<br /> <br /> AM[/itex]