Asked by Mason
A 10kg box is initially at rest 10m above the ground level on top of a hill. The box starts to slide down the hill and when it reaches ground level it has speed 10m/s. Was there a friction force between the box and the ground? If so how much work did the force of friction do as the box moved down the hill?
Answers
Answered by
MathMate
Assuming the first sentence to mean:
"A 10kg box is initially at rest on top of a hill 10m above the ground level."
I.e. the box is resting at the top of the hill, which is 10 m. above ground level below.
Using ground level as zero potential energy, then
At top of hill:
potential energy = mgh = 10*9.8*10 =980 j.
kinetic energy = 0
Total energy = 980 j.
At bottom of hill:
potential energy = mg(0) = 0
kinetic energy = (1/2)mv²
=(1/2)*10*10²
=500 j.
Clearly the box has lost (980-500)=480 j while sliding down the hill. If there is no air resistance, it must have lost tee energy to friction.
"A 10kg box is initially at rest on top of a hill 10m above the ground level."
I.e. the box is resting at the top of the hill, which is 10 m. above ground level below.
Using ground level as zero potential energy, then
At top of hill:
potential energy = mgh = 10*9.8*10 =980 j.
kinetic energy = 0
Total energy = 980 j.
At bottom of hill:
potential energy = mg(0) = 0
kinetic energy = (1/2)mv²
=(1/2)*10*10²
=500 j.
Clearly the box has lost (980-500)=480 j while sliding down the hill. If there is no air resistance, it must have lost tee energy to friction.
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