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. 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

User Romedius
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2 Answers

4 votes

Answer:

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

User Flakes
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5.4k points
3 votes

Answer:


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)

User Notquiteamonad
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