The force Fp = 1200N pulls a box with the mass m=48.0kg along a horizontal track

Question

The force Fp = 1200N pulls a box with the mass m=48.0kg along a horizontal track as shown. The friction coefficent of the track is Uk = 0.360 and the box has the initial velocity Vi = 6.20 m/s. Obtain the friction Force Fk, and the final velocity Vf, if the box is pulled for 20.0m.

(Diagram given)

Fp----> [B1] ---->V1i Fp----> [B2] ----> VF

_________________________________________________

<---Fk [_____ 20m ______] Fk--->

Fk=?

Vf=?

(please an explanation along with the results as to how to solve this would be greatly appreciated!)

Explanation / Answer

frictional force=umg

=0.36*9.8*48

=169.344 N

b)conserving energy,

0.5mv^2=(F-f)s+0.5mu^2

or 0.5*48*v^2=(1200-169.344)*20+0.5*48*6.2^2

or v=29.95 m/s

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