A horizontal spring attached to a wall has a force constant of 890 N/m. A block

Question

A horizontal spring attached to a wall has a force constant of 890 N/m. A block of mass 1.90 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? Score: 1 out of 1 Comment:

(b) What are the two points of interest? Score: 1 out of 1 Comment:

(c) Find the energy stored in the spring when the mass is stretched 5.00 cm from equilibrium and again when the mass passes through equilibrium after being released from rest. x = 5.00 cm J x = 0 cm J

(d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. m/s

(e) What is the speed at the halfway point? m/s Why isn't it half the speed at equilibrium?

A horizontal spring attached to a wall has a force constant of 890 N/m. A block of mass 1.90 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? Score: 1 out of 1 Comment: (b) What are the two points of interest? Score: 1 out of 1 Comment: (c) Find the energy stored in the spring when the mass is stretched 5.00 cm from equilibrium and again when the mass passes through equilibrium after being released from rest. x = 5.00 cm J x = 0 cm J (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. m/s (e) What is the speed at the halfway point? m/s Why isn't it half the speed at equilibrium?

Explanation / Answer

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

A horizontal spring attached to a wall has a force constant of 730 N/m. A block of mass 1.30 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?
1

(b) What are the two points of interest?
2

(c) Find the energy stored in the spring when the mass is stretched 6.80 cm from equilibrium and again when the mass passes through equilibrium after being released from rest.
x = 6.80 3. J
x = 0 4. 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.
5. m/s

(f) What is the speed at the halfway point?
6. 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.)

answer

The spring constant, K = 730 N/m

The mass attached, m = 1.3 kg

c) when, x = 6.8 cm = 0.068 m

Then the energy stored, U = kx^2 = 730(0.068)^2 = 3.38 J

d) According to conservatin of energy the above energy is nothing but the kinetic energy

so, (1/2) mv^2 = 3.38

From the above, v = sqrt(3.38 * 2 / m) = sqrt(3.38 * 2 / 1.3) = 2.28 m/s

So the speed of mass as it passes the equilibrium position is 2.28 m/s

e ) According to conservation of energy we can write

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

       x = 0.068 m and x' = 0 m

      We have

         0.5 k * x^2 = 0.5 m v^2

         0.5 * 730 * 0.068^2 = 0.5 * 1.3 * v^2

         v = 1.6 m/s

f)   half-way point: x=0.068 m; x'=0.034 m

     0.5 * 730 * 0.068^2 = 0.5 * 1.3 * v^2 + 0.5 * 730 * 0.034^2

    ==> v = 1.94 m/s

g) It is a oscillatory motion. So the velocity of the body is slows down at the half

way point and maximum at the equilibrium.

For the parts (a) and (b) I have no idea. Sorry and Thanks

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