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## 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 dont know where i am going wrong... i am also solving part b because part a is theory.

Thank you