The path of motion of a 5 kg collar is y 40 020 x3 At A whe

The path of motion of a 5 kg collar is: y = 4.0 -0.20 x³. At A when x = 0, the collar speed is 4 m/s. Determine the speed of the collar when it reaches B where y = 0. The attached spring has an unstretched length of 2.5 m and stiffness of 225 N/m.

                                                                                                v_B = ______________

Solution

In absence of any external source of energy, the net energy will remain conserved in this case.

That is PE1 + KE1 = PE2 + KE2

Also, the stretch for the spring for position A = 4 - 2.5 = 1.5 m

Stretch at position B = (4/0.2)^1/3 - 2.5 = 0.2144 metres

The energy stored in a spring with stiffness k and elongation, x is given as 0.5kx^2

also, A is at a height of 4 metres above ground, hence it would also have a potential energy of 20g

Hence we get the equation as:

0.5*5*16 + 20*9.81 + 0.5*225*2.25 = 0.5*5*v^2 + 0.5*225*0.2144*0.2144

or, 40 + 196.2 + 253.125 = 2.5v^2 + 5.171328

or, v^2 = 193.66147

or Velocity at point B = 13.9162 m/s