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

Answer:

$v=10\sqrt{2} \ m.s^{-1}$

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:

$KE=\frac{1}{2}\times m.u^2$

$KE=\frac{1}{2} \times 2\times 20^2$

$KE=400\ J$

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

$PE=\frac{1}{2}\times k.\delta x^2$

$PE=\frac{1}{2} \times 100\times 2^2$

$P=200\ J$

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

$400-200=\frac{1}{2} \times m.v^2$

$200=0.5\times 2\times v^2$

$v=10\sqrt{2} \ m.s^{-1}$

Answer 2

Answer:

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

$\frac{1}{2}K\Delta x^{2}=\frac{1}{2}m(v^{2}-u^{2})$

100 x 2 x 2 = 2 x (v² - 400)

200 = v² - 400

v² = 600

v = 24.5 m/s