A sled of weight 195.8 Newtons slipped away from its own from the top of a hill.

Question

A sled of weight 195.8 Newtons slipped away from its own from the top of a hill. The sled loses an altitude of H meters starting from rest to the bottom of the this hill. The thermal energy converted from friction was 2.35 KCal and the sled had been observed to have an initial speed of 30 m/sec at the bottom of the hill. Calculate H the altitude of the hill. Use the principle of conservation of energy for the solution. Calculate the heat of conduction in Watts through 50 cm thick walls steel factory having 6 times the thermal conductivity of glass is 0.042 J/sec.mC. The total area of the walls is 2250

Explanation / Answer

(A) at top :

KE = 0 ( as it starts from rest)

PE = m g H

total energy = m g H = 195.8 H

at bottom,

KE= m v^2 /2 = (195.8/9.8) (30^2)/2 = 8990.8 J
PE = 0

totalenergy = 8990.8 J

energy dissipated = 2.35 kCal = 2.35 x 4184 J = 9832.4 J

Applying energy conservation,

Energy at top = energy at bottom + energy dissipated

195.8 H = 8990.8 + 9832.4

H = 96 m

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