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
AI Thread Summary
To calculate the stopping force for a car weighing 1.5×10^4 N and moving at 45 km/h, the mass is determined to be approximately 1.53e3 kg. The correct method involves using the equation for acceleration, where acceleration can be derived from the initial and final velocities over the stopping distance. An alternative approach utilizes the kinetic energy formula, leading to the equation F = mv^2/2d for calculating the stopping force directly. The discussion emphasizes that both methods are valid, but the energy approach may be more straightforward. Understanding these calculations is essential for accurately determining the stopping force.
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. a = \Delta v/\Delta t \text{ where } \Delta t = 2d/v_i Since final velocity is 0, you have: a = v^2/2d

You are in effect using energy: KE = F*d = \frac{1}{2}mv^2[/tex] which is a little more direct:<br /> <br /> F = mv^2/2d<br /> <br /> AM
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top