# Question with a physics problem

• December 16th 2010, 09:14 AM
colerelm1
Question with a physics problem
A steel mass weighing 0.75kg is released down a frictionless ramp of height 5.0m. The steel mass (m) hits a spring made of copper at the bottom of the ramp. The collision results in both the mass m and the spring (M) to have an increase in temperature from 20.00 degrees Celsius to 20.03 degrees Celsius. How high does the steel mass go after the bounce?

I did some looking around for specific heats and found the specific heat for copper to be 0.385 J/G OC and for steel 500 J kg-1 0C-1.

I have almost no clue how to approach this type of problem, and I would appreciate any help I could get.

I created a quick illustration to help:Attachment 20124
• December 16th 2010, 10:04 AM
snowtea
Use conservation of energy:

Initial potential energy = Final potential energy + Energy lost to heat.

You can find the height from the final potential energy.
• December 18th 2010, 01:50 PM
colerelm1
Ok so I did:

KEi + PEi = KEf + PEf
0 + mgh = .5mv^2 + 0
3 x 9.8 x 5 = .5(3)v^2
solved for v and got 14m/s^2

but now after this what do I do? Did I even do this far correctly? How do I calculate energy lost to heat?
• December 18th 2010, 02:16 PM
snowtea
No.
The initial state is when the block is at the top. You calculated this energy correctly.
The final state when the block rebounds to the top. You also have additional heat at the end.

$
KE_i + PE_i = KE_f + PE_f + Q
$

$
0 + mgh_i = 0 + mgh_f + Q
$

$
Q = MC_{copper} \Delta T_{copper} + mC_{steal} \Delta T_{steal}
$

Where M is mass of spring and m is the mass of the steal mass.
• December 19th 2010, 04:21 PM
colerelm1
so when you say:
Quote:

Originally Posted by snowtea
$
Q = MC_{copper} \Delta T_{copper} + mC_{steal} \Delta T_{steal}
$

Where M is mass of spring and m is the mass of the steal mass.

does that mean the mass multiplied by the change in temperature? I dont think I understand where you're getting
$
Q = MC_{copper} \Delta T_{copper} + mC_{steal} \Delta T_{steal}
$
• December 19th 2010, 04:27 PM
snowtea
$
Q = MC \Delta T
$

This is the heat equation.
Q is heat.
M is mass.
C is specific heat.
Delta T is the change in temperature.

You must have learned this. Otherwise, it would be unreasonable to be assigned this problem.
• December 19th 2010, 04:54 PM
colerelm1
Ok so after solving I end up with:

147 kg x m^2 = 147 kg x m^2 + 49.2750 J

How does this give me an answer though? How will I find how high the steel mass ends up going from this equation?
• December 19th 2010, 05:57 PM
skeeter
Quote:

Originally Posted by snowtea
Use conservation of energy:

Initial potential energy = Final potential energy + Energy lost to heat.

You can find the height from the final potential energy.

read snowtea's initial post again ...
• December 19th 2010, 08:30 PM
colerelm1
ok so the final potential energy is 147 kg x m^2. How do I use that to find the height though?
• December 20th 2010, 05:59 AM
skeeter
Quote:

Originally Posted by colerelm1
ok so the final potential energy is 147 kg x m^2. How do I use that to find the height though?

I have no way of checking your calculations because you have not provided the mass of the copper spring. also, your units for energy are incorrect.

energy in Joules is ...

$\displaystyle J = N \cdot m = \frac{kg \cdot m^2}{s^2}$

here is how to calculate the final height ...

$\displaystyle h_f = \frac{mgh_0 - (Q_c + Q_s)}{mg}$

where $m$ = mass of the steel ball
• December 20th 2010, 07:56 AM
SammyS
Quote:

Originally Posted by colerelm1
Ok so after solving I end up with:

147 kg x m^2 = 147 kg x m^2 + 49.2750 J

How does this give me an answer though? How will I find how high the steel mass ends up going from this equation?

Assuming that $\displaystyle mgh_i=147 \text{ kg}\cdot\text{m}^2/\text{s}^2$ is your initial P.E. (your units were incorrect), and assuming that $\displaystyle 49.2750 \text{ J}$ is the heat energy generated by the collision, you would then have to solve

$\displaystyle 147 \text{ kg}\cdot\text{m}^2/\text{s}^2=mgh_f+ 49.2750 \text{ J}$

for $\displaystyle mgh_f$.

I am curious as to what you used for the mass of the spring.

It looks like you used 3 kg, rather than 0.75 kg, for the mass the steel object.

By the way, $\displaystyle 1 \text{ kg}\cdot\text{m}^2/\text{s}^2=1\text{ J }.$