Question

a particle moves in a straight line with retardation proportional to its displacement its loss of KE for displacement x is proportional to?

Answer 1

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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.

You may get confused why we have used x instead of dx as we defined dx as displacement and x as position. That is because the word ‘displacement’ in the question actually means ‘displacement from starting position’ and not the instantaneous displacement ‘dx’.

In other words -a = kx, where k is a constant of proportionality. Lets substitute this in eqn(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)

which means that

-ΔK ∝ x²

-ΔK is the loss in Kinetic energy.

Hope it helps✌