Calculating Temperature Increase of a Rubber Ball Dropped from a Building Roof

  • Thread starter Thread starter rijo664
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating the temperature increase of a rubber ball after being dropped from a building. The ball, weighing 8.80 kg, is initially propelled down by a spring mechanism storing 987 J of energy and bounces back to a height of 9.5 meters. The participant calculates the gravitational potential energy lost during the fall as 396 J, leading to a total energy of 1383 J when combined with the spring potential energy. There is a query about whether this total energy represents heat energy to determine the temperature change. The conversation also notes the importance of referencing previous discussions for clarity.
rijo664
Messages
25
Reaction score
0
idk if i did this correctly so far so here goes.

The question states

A large rubber ball (8.80 kg) is fired straight down from the roof of a building from a spring loaded mechanism that stored 987 J of elastic potential energy. The building roof is 14.0 meters above ground. The rubber bounces straight back up (noiselessly--don't ask how) back to a height of 9.5 meters. Assuming no heat is lost to the air around the ball, calculate the increase in temperature of the ball (Specific Heat of rubber is 1250 J/kgK

What i did was:
M= 8.80kg
Spring Potential Energy= 987 J
Building roof= 14.0 m
The rubber bounces back up= 9.5 m
Specific Heat of rubber is 1250 J/kgK

1) I found the Change in Height= 14.0-9.5= 4.5m
2) Then i used the gravitational potential energy formula which is
(mass)(g)(the change in height)
3)g=10
4)Gravitational Potential Energy= (8.80)(10)(4.5)
gravitational potential energy= 396 J

That is what i got for my gravitational potential energy then i used my spring potential energy which is 987 J and added that with 396 J and got 1383.
I believe that the 1383 J is the heat and i should find the change in temp. am i right or am i wrong here
 
Last edited:
Physics news on Phys.org
Looks good to me.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top