Homework Help: Net force problem

1. Oct 1, 2011

1. The problem statement, all variables and given/known data

An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 850 kg traveling initially at a speed of 51.0 km/h in a distance equal to the diameter of a dime, which is 1.8 cm? Answer must be in two sig. figs.

2. Relevant equations

V² = u² + 2 a*s
F(net) = m*a

3. The attempt at a solution

51 km/h = 14.16 m/s

0= (14.16)^2 + 2a(0.018)
a= 200.5 / -0.036
a= -5,569.4 m/s^2

850 kg * -5569.4 = Fnet

Fnet = 4,734,027.78 N
(this is 2 sig figs right)?

Did I do this right? If I did, why exactly do I receive -0.036?(I just multiply 0.018 but I don't really understand why.)

Thanks.

2. Oct 1, 2011

Your answer has 2 decimal places. Two sig figs would be $$4.7 * 10^6 N$$. That's probably what they're looking for. You dont' get the negative value from multiplying, that came from solving your equation for "a."

$$V_o^2=-2a_x\Delta x$$

3. Oct 1, 2011

Oh ok thanks. I forgot about that notation.

Ok, even without the negative value, the final answer should still be the same right?(positive)

Last edited: Oct 1, 2011
4. Oct 1, 2011

No, it should be negative because if you're taking the direction of the initial velocity to be in the +x direction (which you are of course) the acceleration is in the opposite (negative) direction because it is bringing the car to a stop. For the force, you wrote:

850 kg * -5569.4 = Fnet

This is correct, but you have to keep the negative when you multiply. You're multiplying a negative and a postive, the result is $$-4.7*10^6 N$$

5. Oct 1, 2011