Momentum and Heat Homework: Mechanical to Thermal Energy

[tamir102]

Homework Statement

A 30 kg boy on a 5 kg sled takes a running start from the top of a slippery hill, moving at 4.00
m/s. The hill is 60 m long at an angle of 11.5 degrees above the horizontal. The boy and sled
reach the bottom of the hill moving at 10.0 m/s. (a) How much mechanical energy has been
transformed into thermal energy? (b) If the steel runners of the sled have a mass of 1.500 kg and
half of the heat generated remains in the runners, by how much does the temperature of the steel
runners rise during the ride downhill? Specific heat for steel = 452 J/(kg·°C). [ANS: 1.94°C]

Homework Equations

Wnc=delta KE+ delta PE
delta T = Q/nc

The Attempt at a Solution

Wnc= delta KE + delta PE
=.5(35)(6^2)+35(9.8)(30)
= 10920 J

delta T = Q/nc
= .5(10920)/(1.5*452)
= 24.16 C


the answer as shown is 1.94 C so i don't know where i am going wrong... i am also solving part b because part a is theory.

Thank you

 

[PeterO]

Homework Statement

A 30 kg boy on a 5 kg sled takes a running start from the top of a slippery hill, moving at 4.00
m/s. The hill is 60 m long at an angle of 11.5 degrees above the horizontal. The boy and sled
reach the bottom of the hill moving at 10.0 m/s. (a) How much mechanical energy has been
transformed into thermal energy? (b) If the steel runners of the sled have a mass of 1.500 kg and
half of the heat generated remains in the runners, by how much does the temperature of the steel
runners rise during the ride downhill? Specific heat for steel = 452 J/(kg·°C). [ANS: 1.94°C]

Homework Equations

Wnc=delta KE+ delta PE
delta T = Q/nc

The Attempt at a Solution

Wnc= delta KE + delta PE
=.5(35)(6^2)+35(9.8)(30)
= 10920 J

delta T = Q/nc
= .5(10920)/(1.5*452)
= 24.16 C


the answer as shown is 1.94 C so i don't know where i am going wrong... i am also solving part b because part a is theory.

Thank you

You can only do part B correctly if you have part A done correctly.
That energy figure is way too high.

For a start .5(35)(10^2) - .5(35)(4^2) is not equal .5(35)(6^2)

That is like saying (a - b)² = a² - b² which it certainly is not

Also the change in height is way less than 30m. The slope is only 11.5^o. The change in height is [co-incidentally] alarmingly close to 11.5 - I thought I had made a calculator error.

 

[tamir102]

you used sine law to find the height correct ?

and part a was just a theoretical answer of explanation not writing.

 

[PeterO]

you used sine law to find the height correct ?

and part a was just a theoretical answer of explanation not writing.

if you call sin(11.5) = h/60 the sine law, then yes.

 

[asteriatic]

for providing your attempt at a solution. Your calculations for part (a) are correct, as the total mechanical energy transformed into thermal energy is 10920 J. However, for part (b), you need to use the specific heat of the steel runners (452 J/(kg·°C)) to calculate the change in temperature. The mass of the steel runners is given as 1.500 kg, so the correct calculation would be:

delta T = Q/(m*nc)
= .5(10920)/(1.500*452)
= 1.94°C

This is because the specific heat of a material is defined as the amount of heat required to raise the temperature of 1 kg of the material by 1°C. Therefore, to find the change in temperature of the steel runners, we need to divide the total heat (Q) by the mass of the runners (m) and the specific heat (nc).

I hope this helps clarify the solution for part (b). Keep up the good work with your studies!