Question

A 2 Kg plain collar C fits snugly on the vertical piston. If the spring is not stretched when the collar is in position 1, determine how quickly the collar is in position 2 (y = 1 m), if (a) is released from rest at 1 and (b ) is released by 1 with an upward speed of 2 m / s. (Solve using the Principle of Conservation of Energy.)

calculate speed in 1

calculate speed in 2

-0,75m ww 3 N/m 2

Answer 1

By conservation of energy, the total amount of kinetic energy gained by is equal to the total potential energy lost.

The equation is:

$\frac 12 m (v_2^2-v_1^2)=mgy-\frac 12 k (x_2-x_1)^2$

$m=2\ kg\\ y=1\ m\\ x_1=0.75\ m\\ x_2=\sqrt{1^2+0.75^2}=1.25\ m\\ k=3\ N/m$

(a) The collar starts from rest, therefore, $v_1=0$

Therefore,

$v_2^2=2\times 9.8-1.5\times 0.25^2=19.50625\\ \implies v_2=4.417\ m/s$

Acceleration on the collar:

$v_2^2-v_1^2=2ay\\ \implies 19.50625=2a\\ \implies a=9.753\ m/s^2$

Therefore, time taken:

$v_2=v_1+at\\ \implies t=4.417/9.753=0.453\ s$

(b)

By energy conservation:

$v_2^2-2^2=2\times 9.8-1.5\times 0.25^2\\ \implies v_2^2=23.50625\\ \implies v_2=4.848\ m/s$

Acceleration on the collar:

$v_2^2-v_1^2=2ay\\ \implies 23.50625-4=2a\\ \implies a=9.753\ m/s^2$

Therefore, time taken:

$v_2=v_1+at\\ \implies t=(2+4.484)/9.753=0.702\ s$