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. Thank you. I am also attaching photos of my incorrect answers.

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 Newton;s second law,
F - f = ma
where f is the frictional force pointing to the oposite direction of F.
a = (F - f ) / m
a = (F - umg) / m
a = (<106,0,0> - <0.15*22*9.8,0,0>) / 22
a = < 73.66, 0, 0> / 22
a = <3.3,0,0> m/s^2
The box will accelerate along the positive x axis.
Its velocity at t=9.6 is from
Vf^2 = V0^2 + a(tf-t0)
Vf = sqrt(V0^2 + a(tf-t0))
Vf = sqrt(6^2 + 2(9.6-8.0) )
Vf = sqrt(39.2)
Vf = 6.26 or
Vf = <6.26,0,0> m/s
It moves along the positive x axis direction a distance d between t=8s and t=9.6s,
d = (1/2)at^2 + V0*t
d = (1/2)*2*(1.5)^2 + 6*1.5
d = 11.25 or
d = <21.25,3,-3> m

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