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

  • Thread starter rijo664
  • Start date
In summary, a rubber ball with a mass of 8.80 kg is fired down from a building roof 14.0 meters above ground using a spring loaded mechanism with 987 J of elastic potential energy. The ball bounces back up to a height of 9.5 meters, and assuming no heat is lost to the air, the increase in temperature of the ball can be calculated using the specific heat of rubber (1250 J/kgK). By finding the change in height and using the gravitational potential energy formula, it can be determined that the gravitational potential energy of the ball is 396 J. When added to the spring potential energy of 987 J, the total energy is 1383 J, which is believed to be
  • #1
rijo664
25
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
  • #2
Looks good to me.
 
  • #3

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

What does "So far am I correct" mean?

The phrase "So far am I correct" is usually used to confirm if one's statement or understanding of a situation is accurate up to a certain point. It is often used in a conversation to check for agreement or to clarify any misunderstandings.

Why do people use the phrase "So far am I correct"?

People use this phrase to make sure that they are on the same page with the person they are talking to. It helps to avoid miscommunication and to ensure that both parties have the same understanding of a situation or topic.

Is the phrase "So far am I correct" commonly used in scientific discussions?

Yes, this phrase is commonly used in scientific discussions as it is important for scientists to have a clear understanding of the information they are discussing. By asking if they are correct so far, it allows for open communication and the opportunity to correct any misunderstandings.

Can the phrase "So far am I correct" be used in formal settings?

Yes, the phrase can be used in formal settings as a way to confirm understanding and to ensure that everyone is on the same page. However, it is important to use it appropriately and respectfully in a formal setting.

Are there any alternative phrases to "So far am I correct"?

Yes, there are alternative phrases that can be used to achieve the same purpose, such as "Am I understanding correctly?" or "Is my understanding accurate so far?" The key is to use a phrase that conveys the same meaning and promotes clear communication.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
31
Views
3K
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
10K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
9K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top