Physics question - Forces on a mass on horizontal surface

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SUMMARY

The discussion centers on calculating the work done on a 2.5 kg mass subjected to a horizontal force of 150 N and a frictional force of 100 N. Participants concluded that without knowing the distance over which the force is applied, it is impossible to determine the work done, as the formula for work (W = F * d) requires both force and distance. The absence of distance or time in the problem statement renders the question incomplete, making it impossible to apply kinematic equations effectively.

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  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of the work-energy principle (W = F * d)
  • Familiarity with kinematic equations
  • Basic concepts of frictional forces
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suzy123
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Physics question -- Forces on a mass on horizontal surface

Homework Statement


A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?


Homework Equations


I'm not too sure if any equation would be used.


The Attempt at a Solution


Again, I am not sure how to answer this question. I think that the work would be 0 but I am not sure.
 
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Hi! Which direction is the horizontal force moving this mass. Can you draw the diagram (if any)?
 
Hi suzy123,

Welcome to Physics Forums!

In addition to the previous comment, there seems to be information missing in the problem statement. Does the original problem statement tell you how far the mass moves, or the distance to which the force is applied? Or maybe the time period that the force is applied? (Anything along those lines?)

Do you know if it is acceptable to give your answer in terms of some other variable like distance traveled or time?
 
Hi there!

This is all of the information i was given for this question and no diagram was included.
 
Different tack: what equations would you have at hand in case we manage to convince you that, in order to come up with a quantitative answer, you do need to make use of some equations nevertheless ?
 
suzy123 said:

Homework Statement


A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?


Homework Equations


I'm not too sure if any equation would be used.


The Attempt at a Solution


Again, I am not sure how to answer this question. I think that the work would be 0 but I am not sure.

suzy123 said:
Hi there!

This is all of the information i was given for this question and no diagram was included.

You need to show some effort of your own, or this thread will be deleted. Our Homework Help rules here on the PF require that you show some effort on your question, and that you do the bulk of the work.
 
I made another attempt at the question.
so I used the f=ma formula. I substituted (150 N-100 N) for f, 2.5 kg for m and I solved for a. I got 20 m/s^2 for a. would I use one of the accelerated motion formulas to possibly solve for d? then substitute it into the w=fd formula?
 
Well I'm not sure how to find the work honestly. Typically the equation w=F*d, however you aren't given distance or time. So I guess you could use (not sure though) the equation v=at. so it seems you have right answer for a, then just equate d=v and you should have a work eq. with t.
 
suzy123 said:

Homework Statement


A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?
The information given are mass,and force.
Even though mass and force gives acceleration,
none of the four kinematics equations will be useful here.
##v=u+at## -
##S=\frac{1}{2}(u+v)t##
##S=ut+\frac{1}{2}at^2##
##v^2=u^2+2as##

The equations require at least time to be given(to find distance) or distance to find work done.
So this question lacks enough information.
It's like saying what is the work done if a force of 1N acts on a mass of 1kg.We don't know the distance the forces acted.It can be 100m,1000m or 10^100m.We don't know.
 

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