Find the amount of energy E dissipated by friction by the time the block stops.

Express your answer in terms of some or all the variables m, v, and h and any appropriate constants.

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]

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

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