Q4.15 Balance (K)
A light horizontal platform is supported at position x=0 by a spring of spring constant k=250 Nm-1. A mass m=5 kg is dropped onto the platform. After a time the oscillations die away and the platform comes to rest at a point x=-a.
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Hint
Reveal
Solution
Reveal
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Solution
- After the oscillations have died away, the mass is in static
equilibrium. Thus
mg=ka; a=mg/k=0.2 m. - (a) positive (movement of mass in direction of gravitational force) ; (b) negative (the mass moves down, hence hfinal<hinitial); (c) negative (mass moves in opposite direction to spring force); (d) positive.
Gravity does work mga=10 J; the gravitational potential energy decreases by 10 J.
The spring force does work
; the potential energy stored by the spring increases by 5 J.
- The total potential energy decreases by 5 J. This energy is dissipated by internal friction in the spring and air drag on the mass as the oscillations die away.