A horizontal spring attached to a wall has a force constant of 820 N/m. A block

Question

A horizontal spring attached to a wall has a force constant of 820 N/m. A block of mass 1.80 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?

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(b) What are the two points of interest?

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(c) Find the energy stored in the spring when the mass is stretched 5.20 cm from equilibrium and again when the mass passes through equilibrium after being released from rest.

(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

a) the system constitutes of the spring and the block.

spring exerts a force towards itself when being strethd and it exerts a force

outwards when compressed.

b)one point of interest is x=0 where spring's potential energy is zero

and total energy in the system=kinetic energy of the block

second point of interest is at the initial amplitude point from where the block started osciallting.

at that point, block's veloicty is zero.

hence total energy in the system=potential energy of the spring

c)when the mass is stretched ,

x=5.2 cm

then energy stored in the spring=0.5*spring constant*x^2

=0.5*820*0.052^2

=1.1086 J

at x=0 , energy =0.5*spring constant*0^2=0 J

d)total energy in the system=1.1086 J

since potential energy of the system at x=0 is 0 J,

if speed of the block is v m/s,

then 0.5*1.8*v^2=1.1086

==>v=sqrt(1.1086/(0.5*1.8))=1.11 m/s

e)at half way point, x=5.2/2=2.6 cm=0.026 m

then spring potential energy=0.5*820*0.026^2=0.27716 J

then kinetic energy of the block=1.1086-0.27716=0.83144 J

so if speed of the block is v,

then 0.5*1.8*v^2=0.83144

==>v=sqrt(0.83144/(0.5*1.8))=0.9612 m/s

it is not half the speed of equilibrium because the energy varies as square of speed and square of compression

rather than linearly.

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