A 16 kg child descends a slide 2 m high and reaches the bottom with a speed of 1

Question

A 16 kg child descends a slide 2 m high and reaches the bottom with a speed of 1.25 m/s. How much thermal energy due to friction was generated in this process?


Known: m=16kg h=2.2m v=1.25m/s
Conservation of Energy
?K+?U+ {changes in all other forms of energy}=0
Ki+Ui+ {changes in all other forms of energy}= Kf+Uf+ {changes in all other forms of energy}
In this problem:
?K+?U+thermal energy=0
?K+?U=-thermal energy
?K=.5m(?v)^2 m=constant
?U=mg?h m & g are constants
Hence, mg?h+.5m(?v)^2=-thermal energy
mg?h=mg(0-2.2)=16*9.8*(-2.2)
.5m(?v)^2=m/2(1.25-0)^2
m/2(1.25-0)^2+16*9.8*(-2.2)=-thermal energy
8(1.25)^2+-344.96=-thermal energy
332.46 J =thermal energy

This is what I have so far. I am not sure of the sign of the final answer . Also could anyone point out any flaws . Other answers I have gotten are 257.46 ,-332.46, and --257.46. Any clarification would be nice.

Explanation / Answer

So by the conservation of energy, Kinetic Energy(KE) should equal to Potential Energy(PE). PE = KE mgh = 0.5*m*v^2 However, if you plug in the number, you will find that mgh > 0.5*m*v^2 The lost of energy is believed to be transferred into thermal energy. PE = KE + TE TE = PE - KE = mgh - .5*mv^2 = 16*9.8*2 - .5*16*1.25^2 = 301.1 J

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