# Work done by and against a force

cnh1995
Homework Helper
Gold Member
Consider a box placed on a smooth surface and a man is pushing it with force F. If the box moves through a distance s, work done "on the block"=F.s and is positive. If the box were moving with some velocity v and the man applied an opposing force F, the block will decelerate and stop. Here also, work is done "on the block by the man" , but F and s are in opposite direction, hence work done is negative here. Whoever applies force, does the work. If a man is moving a body up against gravity such that net force on it is 0, this means mg=Fman.
Since F(man) and s are in the same direction, work done by the man is positive and man loses muscular energy. Work done by gravity is negative since mg and s are opposite in direction. Hence, the body will gain same amount of PE. K.E. of the body will be 0 as discussed earlier( or it is unaffected, if present).

I got a little confused?
You do. Do not think about friction or resistance or reactions yet. If the force makes work and exists anoter force that is directed in opposit to it, that means that both work aganist one to each other. That is much more easily than you think now. Just relax and imagine the arm-wrestling.

Doc Al
Mentor
Also, how is the force the person exerts=mg.
Technically, the gravitational force=mg. If the force the person is applying is equal to mg, then the gravitational force and the person's force would cancel out leaving a net force of zero. So technically, the body shouldn't move?
When the net force is zero, the velocity is constant, not necessarily zero. If you lift the box at constant speed you'll need to exert a force equal to its weight.

When the net force is zero, the velocity is constant, not necessarily zero. If you lift the box at constant speed you'll need to exert a force equal to its weight.
But, as cnh1995 said, for the body to be set into motion, the initial force has to be in infinitesimally greater than mg, just at that instant.. After that, to maintain constant speed, we need to apply a force equal to (mg). Right?

You do. Do not think about friction or resistance or reactions yet. If the force makes work and exists anoter force that is directed in opposit to it, that means that both work aganist one to each other. That is much more easily than you think now. Just relax and imagine the arm-wrestling.
Though work is only done when there is displacement of the body

Consider a box placed on a smooth surface and a man is pushing it with force F. If the box moves through a distance s, work done "on the block"=F.s and is positive. If the box were moving with some velocity v and the man applied an opposing force F, the block will decelerate and stop. Here also, work is done "on the block by the man" , but F and s are in opposite direction, hence work done is negative here. Whoever applies force, does the work. If a man is moving a body up against gravity such that net force on it is 0, this means mg=Fman.
Since F(man) and s are in the same direction, work done by the man is positive and man loses muscular energy. Work done by gravity is negative since mg and s are opposite in direction. Hence, the body will gain same amount of PE. K.E. of the body will be 0 as discussed earlier( or it is unaffected, if present).

Doc Al
Mentor
But, as cnh1995 said, for the body to be set into motion, the initial force has to be in infinitesimally greater than mg, just at that instant.. After that, to maintain constant speed, we need to apply a force equal to (mg). Right?
Sure, if it starts from rest you must first accelerate the box.

cnh1995
Homework Helper
Gold Member
But, as cnh1995 said, for the body to be set into motion, the initial force has to be in infinitesimally greater than mg, just at that instant.
To make KE=0, the force should be infinitesimally larger. Even if the body has nonzero KE and net force is 0, KE is not affected as I've said in earlier post.

Though work is only done when there is displacement of the body
And what? Work is not always exsists even in the presence of motion. It is necessary that present a non-zero projection of force in the direction of the moving. No matter what the sign is. But it does not change anything. The question is simple, and until you complicate it, you will spend the time for nothing. All the best.

Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)

Doc Al
Mentor
Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)
Why not? What if you lower the box?

Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)
The work must be negative aganst the whatever you want force. And it always is! In condition if the "whatever you want force" direction is accepted as possitive. You still think about things that are not actual for you. All should be done in sequence. Let you understand what is work of force against another force is. The other things after that.

cnh1995
Homework Helper
Gold Member
Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)
In my earlier man and box example, when both F and s were in the same direction, work done "by the man" was positive and hence, he lost energy. In the next example also, work done by the man was positive and that by gravity was negative, hence, man lost energy and the body gained gravitational PE. So, by convention, whoever does positive work, loses energy and whoever does negative work, gains energy. If you want to do 'negative' work 'against' gravity, you should gain energy i.e.you should impede the body's motion "due to" gravity. When you catch a free-falling ball, the ball's motion is due to gravity and you impede it, thereby doing negative work against the gravity. Hence, you(your hands actually) gain the energy lost by the ball.

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So, by convention, whoever does positive work, loses energy and whoever does negative work,
That is thermodynamical look on a things. But in mechanical sence it absolutely does not metter what force works against to another.
$$A =F\cdot s\cdot \cos(\alpha )$$
where F and s are modules, but $$cos(\alpha)$$ determines the sign.

Thank you, everyone