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Position update: Initially the bottom of the block is at y = 0.12 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 = 0.07 seconds? y = 1. m

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danialamin

Answer:

The new position is 0.1865 m

Explanation:

As the context of the data is not available, thus following data is utilized from the question as attached above

x_relax=0.32 m

x_stiff=0.13 m

spring stiffness k=9 N/m

mass of block =0.073 kg

t=0.07 s

Velocity of the block is to be estimated thus

Force due to compression in spring is given as

F_s=k Δx

F_s=9(0.32-0.13)

F_s=1.71 N

Force on the block is given as

F_m=mg

F_m=0.073 x 9.8

F_m=0.71 N

Net Force

F=F_s-F_m

F=1.71-0.71 N

F=1 N

As Ft=Δp

So

Δp=1x0.07=0.07 kgm/s

Δp=p_final-p_initial

0.07=p_final-0

p_final=0.07 kgm/s

p_final=m*v_f

v_f=(p_final)/(m)

v_f=0.07/0.073

v_f=0.95 m/s

So now the velocity of the block is 0.95 m/s

time is 0.07 s

y_new=y_initial+y_travel

y_new=0.12+(0.95 x 0.07)

y_new=0.12+0.065

y_new=0.1865 m

So the new position is 0.1865 m

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