Finding Speed to Push Cart up Hill

Click For Summary

Homework Help Overview

The problem involves determining the speed required to push a cart up a hill, given the height of the hill. The scenario includes a toy car moving from a height of 55 cm to a height of 75 cm, and the original poster seeks to understand the necessary speed for the car to reach the second height.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy and kinetic energy in the context of the problem. Questions arise regarding the application of speed in the attempted solution and the role of mass in energy calculations.

Discussion Status

Some participants have provided insights into the energy transformations involved in the problem, suggesting that mass can be treated as a variable that cancels out in the equations. There is an ongoing exploration of how to approach the problem without specific mass values, and multiple interpretations of the energy relationships are being considered.

Contextual Notes

The original poster has an answer sheet indicating a speed of 2 m/s, but there is confusion regarding the mass of the cart and how it affects the calculations. The discussion reflects a lack of consensus on how to proceed with the problem given the constraints of missing information.

ReoFonzo
Messages
23
Reaction score
0

Homework Statement


I have to find the speed needed to push a cart on top of a hill and I only have the height as a number.


Homework Equations


A boy places his matchbox car at the top of the track he has set up with his brother. If the track starts atop the 55 cm toy box, descends to the floor and then up to the 75cm high bed where his brother waits for the car. How fast must the boy push his cart so it just makes it to the brother on the bed ?


The Attempt at a Solution



PE = m(10m/s).55m = 55JM ?:rolleyes:
 
Physics news on Phys.org
You have nothing about speed in your attempted solution. Where do you think it applies?
 
I have an awnser sheet and it says 2m/s... I'm supposed to find what force pushed the cart above the second hill which I assume is kinetic energy but I have no mass which I don't understand...
 
ReoFonzo said:
I have an awnser sheet and it says 2m/s... I'm supposed to find what force pushed the cart above the second hill which I assume is kinetic energy but I have no mass which I don't understand...

Kinetic energy is involved in the problem. The initial speed gives the car initial kinetic energy. During the motion some energy is transformed fom potential to kinetic and then the kinetic is transformed to potential.

The mass is a factor in both the kinetic and potential energies. Its value does not matter. Do the problem with the mass represented by M. In the end you will find the initial speed required is the same for any mass of the car.
 
Last edited:
ReoFonzo said:
I have an awnser sheet and it says 2m/s... I'm supposed to find what force pushed the cart above the second hill which I assume is kinetic energy but I have no mass which I don't understand...

One trick of simple math problems is to immediately start filling things in and use arbitrary numbers if needed, as long as that variable is a part of every term in the equation. This method is particularly useful for ratios between things, where factoring can get complicated. Making up numbers only works if you can look at the equation and think "this can be factored, but i don't feel like doing that."

For the problem given it's as simple as finding the energy difference between where the car starts and where it finishes, which is a difference of 20cm.
(m)(g)(h) = (1/2)(m)(v^2)
m can be factored out, g is gravity (9.something), h is 20cm (convert it to meters)
Solve for v.
 
Thanks a lot guys!
 

Similar threads

  • · Replies 28 ·
Replies
28
Views
3K
Replies
9
Views
4K
Replies
9
Views
5K
Replies
8
Views
4K
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
3K
Replies
1
Views
4K
  • · Replies 12 ·
Replies
12
Views
11K