A particle of mass m moves in a straight line with retardation proportional to its displacement . Find the expression of loss of kinetic energy for any displacement x.
Answers
"Acceleration a = dv/dt,
Velocity v = dx/dt, wherever ‘x’ is position and ‘dx’ is displacement.
Take the magnitude relation of the 2 equations. We get:
a/v = (dv/dt) ÷ (dx/dt)
Cancel the dt, arrange the terms….
vdv = adx —————(i)
The question says that -a ∝ x.
-a = kx, wherever k may be a constant of quotient. Let’s substitute this in equation (i).
-vdv = (k)xdx
Multiply each side with mass m.
-mvdv = (km)xdx
-mvdv = -d(½mv²) = -dK and xdx = d(½x²)
-dK = kilometer × d (½x²)
Integrate each side.
-∫mvdv = ∫ (km)xdx
-ΔK = (km) (½x²)
This means that -ΔK ∝ x²
Therefore, -ΔK is that the loss in K.E..
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@GauravSaxena01
Acceleration a = dv/dt,
Velocity v = dx/dt, where ‘x’ is position and ‘dx’ is displacement.
Take the ratio of the two equations. We get:
a/v = (dv/dt) ÷ (dx/dt)
Cancel the dt, rearrange the terms….
vdv = adx —————(1)
The question says that -a ∝ x.
-a = kx, where k is a constant of proportionality. Let’s substitute this in equation (1).
-vdv = (k)xdx
Multiply both sides with mass m.
-mvdv = (km)xdx
-mvdv = -d(½mv²) = -dK and xdx = d(½x²)
-dK = km * d (½x²)
Integrate both sides.
-∫mvdv = ∫ (km)xdx
-ΔK = (km) (½x²)
This means that -ΔK ∝ x²
Therefore, -ΔK is the loss in Kinetic energy.