Question

. The 2 kg block is given an initial velocity of 20m/s over the smooth rod (i.e. no friction is considered), when it is at A. If the spring has an unstretched length of 1 m and a stiff ness of k = 100 N/m, using conservation of energy determine the velocity of the block, when.s:3 m. g=9.81 nls2

Answer 1

Step-by-step explanation:

mass of block, m = 2 kg

initial velocity, u = 20 m/s

spring constant, K = 100 N/m

elongation, Δx = 3 - 1 = 2 m

Let v be the velocity when s = 3 m.

Use conservation of energy

`(1)/(2)K\Delta x^(2)=(1)/(2)m(v^(2)-u^(2))`

100 x 2 x 2 = 2 x (v² - 400)

200 = v² - 400

v² = 600

v = 24.5 m/s

Answer 2

`v=10√(2) \ m.s^(-1)`

Step-by-step explanation:

Given:

• mass of the block,
  `m=2\ kg`

• initial velocity of the block,
  `u=20\ m.s^(-1)`

• unstretched length of the spring,
  `x=1\ m`

• stiffness constant of the spring,
  `k=100\ N.m^(-1)`

• instantaneous position of the spring,
  `s=3\ m`

Now the change in length of the spring under the weight of the block:

`\delta x=s-x`

`\delta x=3-1`

`\delta x= 2\ m`

Now the kinetic energy of the block when the spring is unstretched:

`KE=(1)/(2)* m.u^2`

`KE=(1)/(2) * 2* 20^2`

`KE=400\ J`

Now the spring potential energy of the spring when it is compressed:

`PE=(1)/(2)* k.\delta x^2`

`PE=(1)/(2) * 100* 2^2`

`P=200\ J`

Now the remaining part of the kinetic energy initially possessed by the body is the instantaneous kinetic energy of the block:

`\Delta E=KE_f`

where:

`\Delta E=` change in energy

`KE_f=` final kinetic energy at the given instant

`400-200=(1)/(2) * m.v^2`

`200=0.5* 2* v^2`

`v=10√(2) \ m.s^(-1)`