Net Force Problem: Stopping an Auto in 1.8 cm

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Homework Help Overview

The problem involves calculating the net force required to stop an automobile of mass 850 kg traveling at a speed of 51.0 km/h over a distance of 1.8 cm. The context is rooted in kinematics and dynamics, specifically focusing on concepts of acceleration and force.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find acceleration and subsequently the net force. They question their calculation of a negative value in their results and seek clarification on the significance of this value.

Discussion Status

Participants are engaged in clarifying the calculations and the implications of negative values in the context of direction. Some guidance has been provided regarding the interpretation of acceleration and force directionality, but there is no explicit consensus on the final interpretation of the results.

Contextual Notes

Participants are discussing the significance of significant figures in their final answer and the implications of direction in their calculations. The problem setup involves specific constraints, such as the distance being very small (1.8 cm), which may influence the interpretation of results.

Crusaderking1
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Homework Statement



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.



Homework Equations



V² = u² + 2 a*s
F(net) = m*a


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

[tex]V_o^2=-2a_x\Delta x[/tex]
 
AdkinsJr said:
Your answer has 2 decimal places. Two sig figs would be [tex]4.7 * 10^6 N[/tex]. That's probably what they're looking for. You dont' get the negative value from multiplying, that came from solving your equation for "a."

[tex]V_o^2=-2a_x\Delta x[/tex]

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:
Crusaderking1 said:
Oh ok thanks. I forgot about that notation.

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

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 positive, the result is [tex]-4.7*10^6 N[/tex]
 
AdkinsJr said:
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 positive, the result is [tex]-4.7*10^6 N[/tex]

Thanks for the clarification. Much appreciated.
 

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