The question: An 8 kg block with a speed of 2 m/s begins to be pulled for5 meter

Question

The question:

An 8 kg block with a speed of 2 m/s begins to be pulled for5 meters horizontally across a floor with a force of 45 N as shown.The coefficient of kinetic friction on the surface is 0.2.

b) How much work is done by the gravitational force?
c) How much work is done by the normal force?

As per my professor's solution to this problem, (for b)

Wmg = F·d
Wmg = (mg)(5)(cos 90) = 0 J

(for c)

WFn = F·d
WFn = (FN)(5)(cos 90) = 0 J

I understand that the normal force is a surface'scounter against the force being applied onto by the objectingresting on it, thus the work done by the normal force would be thesame as the work done by the gravitational force. I just don't getwhere cos 90o came from! I would really appreciate ifanyone could explain to me where cos 90o came from.Thanks!


Wmg = F·d
Wmg = (mg)(5)(cos 90) = 0 J

(for c)

WFn = F·d
WFn = (FN)(5)(cos 90) = 0 J

I understand that the normal force is a surface'scounter against the force being applied onto by the objectingresting on it, thus the work done by the normal force would be thesame as the work done by the gravitational force. I just don't getwhere cos 90o came from! I would really appreciate ifanyone could explain to me where cos 90o came from.Thanks!


I understand that the normal force is a surface'scounter against the force being applied onto by the objectingresting on it, thus the work done by the normal force would be thesame as the work done by the gravitational force. I just don't getwhere cos 90o came from! I would really appreciate ifanyone could explain to me where cos 90o came from.Thanks!

Explanation / Answer

You know W=f*d, where f and d are parallel. when we have anglebetween f and d we hve to find parallel component usually it isfcos. the angle between normal force, gravity and d is 90because they are perpendicular and cos90=0 if we have f and d parallel (the angle is 0 and cos0=1) we canwrite only f*d Sorry for my english

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