Magnitude of the additional force

[Pose]

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A 3kg block slides with a constant velocity at 3 m/s along a flat surface with a coefficient of kinetic friction = 0.5. At a certain point, an additional force is applied, and the block travels 44m in 4 seconds. What is the magnitude of the additional force?

This seems so easy, and yet I don't get it. I tried to do this:

d = 44m
t = 4 s
vi = 3
vf = 11
a = ?

vf = vi + at --> 11 = 3 + a4 --> a = 2 m/s^2

F = ma = 6 N.

I also tried to do something weird with W = Fd = 1/2mv^2...that got me nowhere.

Answer is 12 N.

 

[mehc012]

A 3kg block slides with a constant velocity at 3 m/s along a flat surface with a coefficient of kinetic friction = 0.5. At a certain point, an additional force is applied, and the block travels 44m in 4 seconds. What is the magnitude of the additional force?

This seems so easy, and yet I don't get it. I tried to do this:

d = 44m
t = 4 s
vi = 3
vf = 11
a = ?

vf = vi + at --> 11 = 3 + a4 --> a = 2 m/s^2

F = ma = 6 N.

I also tried to do something weird with W = Fd = 1/2mv^2...that got me nowhere.

Answer is 12 N.

You're ignoring some of the variables you wrote down!

Go with:

x_f = x₀ + v₀t + ½at²

44m = 4s·3m/s + ½a·16s²
32m/s² = 8a
a = 4m/s²

F = 3kg·4m/s²
F = 12N

The reason your method did not work was that your calculation of v_f was incorrect...what you calculated was the average velocity, not the final velocity.

 

[Pose]

WOW, I amaze myself sometimes. Thanks for pointing that out!

 

[mehc012]

wow, i amaze myself sometimes. Thanks for pointing that out!

PS, that is the grin smiley, but for some reason SDN keeps changing it. 😕