Elementary - braking force of a trolley

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Homework Help Overview

The problem involves calculating the braking force required for a trolley weighing 16,000 kg to come to a stop over a distance of 10 m, starting from an initial velocity of 6 m/s. The discussion centers around the relationships between force, mass, acceleration, and distance in the context of motion and energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between force, mass, and acceleration, questioning how to solve for two unknowns (force and time) with limited equations. Some suggest using kinematic equations to relate acceleration and distance, while others reference the work-energy principle.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to the problem. Some have provided equations and attempted to manipulate them, while others express confusion about the results and seek clarification on their reasoning. There is no clear consensus yet on the correct approach or solution.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a noted uncertainty regarding the assumptions made in the calculations.

Glyper
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Homework Statement



A trolley weighing 16 000kg is moving along a horizontal surface with a velocity od 6m/s. What braking force would cause it to brake on a distance of 10m and how long it would take?

Homework Equations



a = F/m = ΔV/t

The Attempt at a Solution



So ΔV = 6m/s, m=16 000kg, s = 10m and F = mΔV/Δt. We know neither F nor t so how could I solve this? It's just one equation but two unknown variables. Could you please help?
 
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Do you know the relationships between the work done by a force acting on a body and its energy?
 
I know Newton's laws of motion, that's why I have the a=F/m, but what does that give me?
 
If a, acceleration, is constant, then v= at+ v0 and d= (1/2)at^2+ v0t.

You know that v0= 6 m/s and that d= 10m . Solve the quadratic equation (1/2)at^2+ 6t- 10= 0 for t, with parameter a. The final velocity was 0 so at+ 6= 0. Put the result you got for t into that and solve the resulting equation for a.
 
Rather have a look at the chapter on work and energy.
 
Do you know what the answer should be?
 
HallsofIvy - OK, thank you. So we get t = (2 (sqrt(5 a+9)-3))/a (the other one is negative so doesn't matter), we plug it into the latter and get (2 (sqrt(5 a+9)-3) so (sqrt(5 a+9)=3 but then... a would have to be 0, which it's not, is it? Where did I make a mistake?
 
Could someone please tell me where did I make a mistake? :(
 
No matter what I do to the expression, it never outputs anything other than 0 - why? Do I lack some variable?
 

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