A horizontal spring attached to a wall has a force constant of 740 N/m. A block

Question

A horizontal spring attached to a wall has a force constant of 740 N/m. A block of mass 1.00 kg is attached to the spring and oscillates freely on a horizontal, frictionless surface as in the figure below. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released.

a) What objects constitute the system, and through what forces do they interact?


(b) What are the two points of interest?


(c) Find the energy stored in the spring when the mass is stretched 5.40 cm from equilibrium and again when the mass passes through equilibrium after being released from rest.

x = 5.40

  J

x = 0

  J

(d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. (Do this on paper. Your instructor may ask you to turn in this work.)

(e) Substitute to obtain a numerical value.
  m/s

(f) What is the speed at the halfway point?
  m/s

(g) Why isn't it half the speed at equilibrium? (Do this on paper. Your instructor may ask you to turn in this work.)

x = 5.40

  J

x = 0

  J

Explanation / Answer

Using law of conservation of energy

Energy at the Equilibrium point = Energy at the extreme point

0.5*m*v^2 = 0.5*k*x^2

m is the mass of the body

v is the speed of the body at equilibrium

k is the spring constant

x is the compression of the spring = 5.4 cm = 0.054

m*v^2 = k*x^2

1*v^2 = 740*0.054^2

v = 1.46 m/sec

a) spring and block are constituted as system

spring force ,Normal force and gravitational forces are teh interacting forces

b) Equilibrium point and two expreme points are the points of intrest

C) at x = 5.4 cm

E = 0.5*k*x^2 = 0.5*740*0.054^2 = 1.078 J

at x= 0 m

E = 1.078 J

D) sing law of conservation of energy

Energy at the Equilibrium point = Energy at the extreme point

0.5*m*v^2 = 0.5*k*x^2

m is the mass of the body

v is the speed of the body at equilibrium

k is the spring constant

x is the compression of the spring = 5.4 cm = 0.054

m*v^2 = k*x^2

v = x*sqrt(k/m)

E) v = x*sqrt(k/m) = 0.054*sqrt(740/1) = 1.47 m/sec

f) using law of conservation of energy

energy at the equilibrium point = energy at the halfway point

0.5*m*v^2 = (0.5*m*u^2)+(0.5*k*x^2)

0.5*1*1.47^2 = (0.5*1*u^2)+(0.5*740*(0.054/2)^2)

u = 1.27 m/sec

g) here it was acted by a spring force which changes with distance from the equilibrium point

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