Question

A spring has a relaxed length of 35 cm (0.35 m) and its spring stiffness is 12 N/m. You glue a 66 gram block (0.066 kg) to the top of the spring, and push the block down, compressing the spring so its total length is 16 cm. You make sure the block is at rest, then at time t = 0 you quickly move your hand away. The block begins to move upward, because the upward force on the block by the spring is greater than the downward force on the block by the Earth. Calculate approximately y vs. time for the block during a 0.24-second interval after you release the block, by applying the Momentum Principle in three steps each of 0.08-second duration.
Step 1

Force: Just after releasing the block, calculate the force exerted on the block by the spring, the force exerted on the block by the Earth, and the net force: Fspring, y = ? FEarth, y = ? Fnet, y = ?.

Momentum update: Just after releasing the block, the momentum of the block is zero. Approximate the average net force during the next time interval by the force you just calculated. At t = 0.08 seconds, what will the new momentum and velocity of the block be?

Position update: Initially the bottom of the block is at y = 0.13 m. Approximating the average velocity in the first time interval by the final velocity, what will be the new position of the bottom of the block at time t = 0.08 seconds?
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Step 2

Force: At the new position, calculate the force exerted on the block by the spring, the force exerted on the block by the Earth, and the net force: Fspring, y = N FEarth, y = N Fnet, y = N.

Momentum update: Approximate the average net force during the next time interval by the force you just calculated. At time t = 2.

Answer 1

We have that the value for y is

$y=0.12376m$

 

From the Question we are told that

Relaxed length of 35 cm (0.35 m)

Spring stiffness is 12 N/m.

Glue a 66 gram block (0.066 kg)

Total length is l_t=16 cm

Block during a 0.24-second interval

Generally the equation for the Force of spring of block  is mathematically given as

$F_{spring}=kx\\\\\ F_{spring}=12*(0.35-0.16)\\\\ F_{spring}=2.28N$

Generally

$F_{elastic}=-mg\\\\F_{elastic}=-0.066*9.8\\\\F_{elastic}=-0.6468N$

Generally,Net Force

$F_{net}=F_{spring}-F_{elastic}\\\\F_{net}=2.28N-(-0.6468N)\\\\F_{net}=2.9268N$

Generally,Velocity of block

$V_y=\frac{py}{m}$

Where

$Py= F_{net}*dt\\\\Py= 2.9268*0.08\\\\Py=0.234144$

Therefore

$V_y=\frac{0.234144}{0.066}\\\\V_y=3.547m/s$

Generally the equation for the Velocity of block   is mathematically given as

$Vy=\frac{yf-yt}{t-t_0}\\\\y=(3.547(0.08-0))-0.16$

$y=0.12376m$

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