A 22 kg box is being pushed across the floor by a constant force ‹ 106, 0, 0 › N

Question

A 22 kg box is being pushed across the floor by a constant force ‹ 106, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.15. At t = 8.0 s the box is at location ‹ 13, 3, ?3 › m, traveling with velocity ‹ 6, 0, 0 › m/s. What is its position and velocity at t = 9.6 s? - Please show work and explain. I am trying to learn :) Thank you.

A 22 kg box is being pushed across the floor by a constant force N. The coefficient of kinetic friction for the table and box is 0.15. At t = 8.0 s the box is at location m, traveling with velocity 6, 0, 0 > m/s. what is its position and velocity at t = 9.6 s? New velocity v= m/s New position d =

Explanation / Answer

from the given data

location of initual position,

xo = 13 m

yo = 3 m

zo = -3 m

vox = 6 m/s

voy = 0

voz = 0

Fx = 106 N

Fy = 0

Fz = 0

from the given data,

ax = Fx/m

= 106/22

= 4.82 m/s

x = xo + vox*t + 0.5*ax*t^2

= 13 + 6*9.6 + 0.5*4.82*9.6^2

= 292.7 m

so, d = <292.7,3,-3> m <<<<<<<<<---------------Answer

at t = 9.6 s

vx = vo + ax*t

= 6 + 4.82*9.6

= 52.25 m/s

vy = 0

vz = 0

so, v = <52.25, 0 , 0> m/s <<<<<<<<<---------------Answer

since the force is in only x direction coardinate of x and x-component of velcoity only change.

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