2. A spring of negligible mass and force constant 420 N/m is hung vertically, an

Question

2. A spring of negligible mass and force constant 420 N/m is hung vertically, and a 0.200-kg pan is suspended from its
lower end. A butcher drops a 2.50-kg steak onto the pan, from a height of 0.25 m (as measured from the pan). The
steak undergoes a complete inelastic collision and the system oscillates periodically.
(i) What is the speed of the steak and pan after the collision?

(ii) With the steak on the pan, the system has a new equilibrium point. Use F = ma to solve for the new equilibrium
point, and give the starting position of the pan using the equilibrium point as zero.
(iii) What is the amplitude of the oscillation?
(iv) What is the period of the motion?

(v) Is the contact force ever equal to zero? If so calculate the position where that is the case.

Explanation / Answer

1)

moments are conserved, when the steak hits the pan:

velocity v of the steak:

v = ?(2gh) = ?(2*9.81*0.25) = 2.215 m/s

velocity pan + steak is

v = m1v1/(m1+m2) = 2.5*2.215/(2.7) = 2.050 m/s

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2)

old equilibrium point: x = F/k = 0.2*9.81/420 = 0.00467 m below the end of the relaxed spring

new equilibrium point: x = F/k = 2.7*9.81/420 = 0.063 m below the end of the relaxed spring.

let the motion start at the new equilibrium point. In fact it starts 6 cm above this point.

---------

3).

initial kinetic energy of pan + steak = spring energy

1/2 mv^2 = 1/2 kA^2 with A = amplitude

2.7* 2.05^2 = 420*A^2

A = 0.1643 cm ---> amplitude

------

4)

T = 2pi?(m/k) = 2pi?(2.7/420) = 0.503 s

-----

5

yes, at the highest point of the movement (top dead center)

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