Elementary - braking force of a trolley

In summary, the conversation is discussing a problem involving a trolley with a weight of 16,000kg moving at 6m/s and needing to brake on a distance of 10m. The homework equations used are a=F/m and a=F/m=ΔV/t. However, there is only one equation and two unknown variables (F and t). To solve this, the relationships between work done by a force and energy are discussed. The final velocity of the trolley is 0, so the quadratic equation (1/2)at^2+6t-10=0 is used to solve for t, with parameter a. The answer should be t = (2 (sqrt(5 a+9)-3
  • #1
Glyper
27
0

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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  • #2
Do you know the relationships between the work done by a force acting on a body and its energy?
 
  • #3
I know Newton's laws of motion, that's why I have the a=F/m, but what does that give me?
 
  • #4
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.
 
  • #5
Rather have a look at the chapter on work and energy.
 
  • #6
Do you know what the answer should be?
 
  • #7
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?
 
  • #8
Could someone please tell me where did I make a mistake? :(
 
  • #9
No matter what I do to the expression, it never outputs anything other than 0 - why? Do I lack some variable?
 

1. What is the braking force of a trolley?

The braking force of a trolley refers to the force required to slow down or stop the trolley's motion. It is typically measured in Newtons (N) and is affected by various factors such as the mass of the trolley, the speed of the trolley, and the surface on which it is traveling.

2. How is the braking force of a trolley calculated?

The braking force of a trolley can be calculated using the formula F = m x a, where F is the force, m is the mass of the trolley, and a is the acceleration. This means that the braking force is directly proportional to the mass of the trolley and the rate at which it is decelerating.

3. What factors affect the braking force of a trolley?

The braking force of a trolley is affected by several factors, including the mass of the trolley, the speed at which it is traveling, the surface it is traveling on, and the presence of any external forces such as friction or air resistance. In general, the heavier the trolley and the faster it is moving, the greater the braking force required to slow it down.

4. How can the braking force of a trolley be increased?

The braking force of a trolley can be increased by increasing the friction between the trolley and the surface it is traveling on. This can be achieved by using materials with higher coefficients of friction, such as rougher surfaces or by adding brakes or other devices to increase the frictional force acting on the trolley.

5. Why is it important to understand the braking force of a trolley?

Understanding the braking force of a trolley is important in various fields such as engineering, transportation, and physics. It allows us to design and optimize braking systems for various types of trolleys, ensuring their safe and efficient operation. It also helps us understand the principles of motion and forces in physics and how they apply to real-world scenarios.

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