Intro Physics -- Cart launched by a spring slides up a ramp...

Click For Summary
SUMMARY

The discussion centers on calculating the maximum height a 5 kg cart, launched with an initial velocity of 12.3 m/s from a spring, reaches on a 4 m ramp. The kinetic energy (Ek) is calculated as 378.225 J using the equation Ek = 1/2 mv². The gravitational potential energy (Eg) at the ramp height is 196 J, leading to a total mechanical energy (Emech) of 574.225 J. The forum participants emphasize the importance of understanding the relationship between kinetic and potential energy, as well as the correct application of equations of motion to determine the distance traveled along the ramp.

PREREQUISITES
  • Understanding of kinetic energy (Ek = 1/2 mv²)
  • Knowledge of gravitational potential energy (Eg = mgh)
  • Familiarity with the conservation of mechanical energy principle
  • Basic equations of motion, particularly vf² = vi² + 2ad
NEXT STEPS
  • Study the conservation of mechanical energy in physics problems
  • Learn how to apply equations of motion in non-linear scenarios
  • Explore potential energy concepts and their independence from the path taken
  • Practice solving physics problems involving energy transformations
USEFUL FOR

Students studying introductory physics, particularly those focusing on energy conservation and motion dynamics, as well as educators seeking to enhance problem-solving techniques in physics education.

zan123_
Messages
7
Reaction score
0
New user has been reminded to always show their work on schoolwork problems.
Homework Statement
The cart is set up with a frictionless apparatus as shown in the figure below. If the cart launched from the spring has an initial velocity of 12.3m/s how high up the ramp would it go? Show your work.
Relevant Equations
The mass of the cart is 5 kg. The ramp is 4 m. Pls let me know if you need more info
javalab20230725073519.png
 
Physics news on Phys.org
zan123_ said:
Homework Statement: The cart is set up with a frictionless apparatus as shown in the figure below. If the cart launched from the spring has an initial velocity of 12.3m/s how high up the ramp would it go? Show your work.
Relevant Equations: The mass of the cart is 5 kg. The ramp is 4 m. Pls let me know if you need more info

View attachment 329629
What is your assessment so far?
 
could you clarify what do you mean? it is on a 5 kg cart moving down a 4 m ramp. i don't know what equations i should use to calculate this
 
zan123_ said:
could you clarify what do you mean? it is on a 5 kg cart moving down a 4 m ramp. i don't know what equations i should use to calculate this
What standard equations or conservation laws have you been taught that might relate to this?
On this forum, you are required to show some attempt.
 
oh okay, i have learned equations but i have a hard time applying them in word questions like these. we dont have a lot of problems to practise so i've been having trouble. Since I have the velocity would i do:

Ek=1/2mv^2
= 1/2(5)(12.3)^2
Ek=378.225 J

Eg=mgh
=(5)(9.8)(4)
196 J

Emech=ek+eg
=574.225 J

i honestly dont know how to find the distance
 
zan123_ said:
i honestly dont know how to find the distance
Consider the following:
What is the velocity of the cart when it is at the distance you're looking for?
Play a movie in your head that shows the motion of the cart from the moment it starts moving until it reaches the required distance. Then think how this motion could be related to the equations you posted in #5.
 
Last edited:
does this work?

vf^2=vi^2 + 2ad
0 = 12.3^2 +2(9.8)d
0 = 151.29 + 19.6d
d = 15129/19.6
d=7.7 m
 
It doesn't work because it assumes that the acceleration of the cart is the acceleration of gravity, i.e. that the cart is in free fall which it isn't.
 
  • Like
Likes   Reactions: PeroK
kuruman said:
It doesn't work because it assumes that the acceleration of the cart is the acceleration of gravity, i.e. that the cart is in free fall which it isn't.
However, the approach only needs a tiny bit of additional reasoning in order to become correct. Something about how potential energy is independent of the path.
 
  • #10
jbriggs444 said:
However, the approach only needs a tiny bit of additional reasoning in order to become correct. Something about how potential energy is independent of the path.
I think the additional reasoning should also include the answer to my question
kuruman said:
What is the velocity of the cart when it is at the distance you're looking for?
the significance of which I am not convinced that OP understands.
 

Similar threads

Replies
18
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
14
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K