Respuesta :
Answer:
The amount of energy E dissipated by friction by the time the block stops is given by
[(1/2)(m)(v²) + mgh]
Explanation:
Using the work-energy theorem,
The workdone in moving a body from one point to another is equal to the change in kinetic energy of the body between those two points.
ΔK.E = W
ΔK.E = (final kinetic energy) - (Initial kinetic energy)
(final kinetic energy) = 0 J (This is because the body comes to rest at the end)
(Initial kinetic energy) = (1/2)(m)(v²)
where m = mass of the body
v = initial velocity of the body
ΔK.E = 0 - (1/2)(m)(v²)
ΔK.E = -(1/2)(m)(v²)
Workdone between the start of the motion and the end = (Workdone by frictional force all through the motion) + (Workdone by gravity in descending the ramp)
Let the work done by Friction be Wf
Workdone by gravity in descending the ramp = (- mg) (- h) = mgh
where h = vertical height descended
g = acceleration due to gravity
ΔK.E = Wf + mgh
Wf = ΔK.E - mgh
Wf = -(1/2)(m)(v²) - mgh
Wf = - [(1/2)(m)(v²) + mgh]
The amount of energy E dissipated by friction by the time the block stops = magnitude of the work done by frictional forces = [(1/2)(m)(v²) + mgh]
The amount of energy E dissipated by friction by the time the block stops is 1/2mv² + mgh.
What is Energy?
Thus can be defined as the ability or capacity to do work.
Using the work-energy theorem,
The work done in moving a body from one point to another is Kinetic energy.
ΔK.E = W
ΔK.E = (final kinetic energy) - (Initial kinetic energy)
1/2mv² - 0 J = 1/2mv²
Work done between the start and end of motion = Work done by frictional force + Work done by gravity.
Let the work done by Friction be Wf
Work done by gravity = mgh
where is mass, g is acceleration due to gravity , h = vertical height.
-ΔK.E = Wf + mgh
Wf = -ΔK.E - mgh
Wf = 1/2mv² + mgh
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