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

In summary: So in summary, the cart would go up 7.7m if launched from the spring with an initial velocity of 12.3m/s.
  • #1
zan123_
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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
 
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  • #2
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?
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
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:
  • #7
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
 
  • #8
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.
 
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  • #9
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.
 

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

What is the principle behind the motion of the cart launched by a spring?

The motion of the cart launched by a spring is primarily governed by the conservation of energy. Initially, the spring stores potential energy when compressed or stretched. When released, this potential energy is converted into kinetic energy, propelling the cart. As the cart moves up the ramp, its kinetic energy is converted into gravitational potential energy until it comes to a stop.

How do you calculate the initial velocity of the cart?

The initial velocity of the cart can be calculated using the conservation of energy principle. The potential energy stored in the spring is given by \( \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the compression/stretch distance. This energy is converted into the kinetic energy of the cart, which is \( \frac{1}{2} m v^2 \), where \( m \) is the mass of the cart and \( v \) is its velocity. Setting these equal gives \( v = \sqrt{\frac{k x^2}{m}} \).

What factors affect the distance the cart travels up the ramp?

The distance the cart travels up the ramp is influenced by several factors: the initial kinetic energy (which depends on the spring constant and compression/stretch distance), the angle of the ramp, the mass of the cart, and frictional forces. The steeper the ramp or the greater the friction, the shorter the distance the cart will travel.

How do you account for friction in the calculations?

To account for friction, you need to include the work done against friction in the energy conservation equations. The work done by friction is \( f d \), where \( f \) is the frictional force and \( d \) is the distance traveled. This work reduces the mechanical energy of the system, so the equation becomes \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 + f d \) for the initial motion and \( \frac{1}{2} m v^2 = m g h + f d \) for the motion up the ramp, where \( g \) is the acceleration due to gravity and \( h \) is the height gained.

How can you experimentally determine the spring constant?

The spring constant \( k \) can be experimentally determined using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement: \( F = k x \). By measuring the force required to compress or stretch the spring by a known distance, you can calculate \( k \). This can be done using a force sensor or by hanging known masses and measuring the displacement they cause.

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