Work and distance

[qw098]

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Quick question I have tough time understanding if someone can help please!

A child must traverse a certain distance over the rough surface at the base of a downward sloping slide before coming to a complete stop. If the slide designers wished to stop the child in half that distance, the work required by friction would necessarily:

A) double
B) quadruple
C) remain the same
D) halve

The answer is C). (<-- Highlight to see correct answer)

I was wondering why! I chose the answer D) (<-- Highlight to see correct answer) because I thought if I apply less work the distance will decrease because of the equation W=Fxd!

Thanks!

 

[Charles Darwin]

W = F * D => If you halve the distance, you must double the force (in this case, friction). The work would remain the same.

 

[Jepstein30]

Work is unaffected by machines. This is a common trick question.

If they asked about the force, however, that would have to double.

 

[qw098]

Thanks guys!

So.. when does work ACTUALLY change!? (How can I increase or decrease work?) I guess I could use the W=PE formula and increase height to increase work?

 

[Charles Darwin]

Thanks guys!

So.. when does work ACTUALLY change!? (How can I increase or decrease work?) I guess I could use the W=PE formula and increase height to increase work?

Change mass, velocity, height

 

[milski]

The work here is to "dissipate" the kinetic energy of the child. That depends only on the mass of the child and its velocity at the end of the slide.

 

[ebasappa]

Just to make sure I'm reasoning correctly, from what I understand it remains the same because in both cases you start with the same KE. Work is dissipating that energy, so it'll be the same either way. If it asked for FORCE, then you would have to double it because W=F(D), and D=D/2. Right?...

 

[Charles Darwin]

Just to make sure I'm reasoning correctly, from what I understand it remains the same because in both cases you start with the same KE. Work is dissipating that energy, so it'll be the same either way. If it asked for FORCE, then you would have to double it because W=F(D), and D=D/2. Right?...

👍

 

[Singhp03]

Work is unaffected by machines. This is a common trick question.

If they asked about the force, however, that would have to double.

Hate these problems, I nearly always fall for it!! uhhh But your definitely right, work is unaffected, the engineers would just have to apply more force to stop the child: