A horizontal spring attached to a wall has a force constant of 760 N/m. A block

Question

A horizontal spring attached to a wall has a force constant of 760 N/m. A block of mass 1.70 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? This answer has not been graded yet. (b) What are the two points of interest? Score: 0 out of 0 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 Correct: Your answer is correct. J x = 0 cm Correct: Your answer is correct. 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. Correct: Your answer is correct. m/s

(e) What is the speed at the halfway point?

I need help calculating (e) I was able to calculate the other answers.

Explanation / Answer

here,
k= 760 N/m
mass of block = 1.70 Kg

The system is
1.) the block
2.) the spring
3.) the wall
4.) the frictionless surface.

The block interacts with the spring through the spring's force due to compression or stretching .

The spring interacts with the wall through a normal force since the spring does not either crash through the wall nor pull off it.

The block interacts with gravity and it also acts with the frictionless surface through the normal force.

The Two point of intrest are the extremist of compression and streteching,

Now,
Total Displacement = X = 5 cm = 0.05 m

Potential Energy when strtched = PE = 0.5 * k * x^2
PE = 0.5 * 760 *0.05^2
PE = 0.95 J

Total Displacement = X = 0
Potential Energy = PE = 0.5 * k * 0
PE = 0

the speed at equillibrium is found using conservation of mechanical energy:
1/2mv^2 = 1/2kx^2

where x = 0.05 m
v = sqrt(k/m)*x
v = sqrt(760/1.70)*0.05
v = 1.05 m/s

At Halfway
x = 0.05 / 2 = 0.025 m
Now,

Total Energy = 0.95 J
0.5 * m * v^2 + 0.5 k x^2 = 0.95
0.5 * 1.7 * v^2 + 0.5 * 760 * 0.025^2 = 0.95
v = sqrt( ( 1.9 - 0.23 ) / 1.7 )
v = 0.91 m/s

At halfway the velocity wil be equal to 0.91 m/s

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