This is kind of confusing to me. I'm guessing that it's specific to the problem. Is the work done by friction always negative? Is the work done by gravity always negative? Spring as well?
It seems like work done by friction should be negative, but then there was a problem that said "Determine the increase in internal energy of the crate–incline system owing to friction." Basically, the answer was a positive work in friction. It's confusing to me.
Whenever a given force and displacement produced by that force has an angle >90 degrees,the work done is said to be negative. Since by the definition of work done,W= F.s = Fs cos(angle between f and s),thus whenever angle is >90, cos(angle) is negative and hence work done is negative.
Coming on your question Is the work done by friction always negative ,my answer is no.Consider a rolling tyre,the direction of friction on the tyre is same as direction of motion so here the work done by friction is positive.
Is the work done by gravity always negative The answer again is no,when a body is falling down, the direction of gravitational force and motion is same so work done is positive.
Is the work done by spring force is always negative The answer is again no,when you stretch/compress a spring the work done by spring force is negetive but when you release the stretched/compressed spring work done by spring force is positive.
All you have to do is to identify direction of force acting and the direction in which body is displaced.
Force is a vector, meaning magnitude and direction. Work done by a force is relative to the direction of a force is the scalar value obtained by performing the vector dot product of the force and the displacement (which is also a vector). If something isn't coming out to what you expect when you compute work, make sure you have the right magnitude and direction for everything. The direction of force is not always intuitive.
In the case of a displacement that does not have a uniform force along its length, you would have to use the calculus integral to compute the scalar work done by said force.
The reason work can be negative is because it's possible for something to move counter to one of the forces that is exerted on it (because another force overpowered it or it was already moving in such a way the force didn't get a chance to overpower it). In fact work can be negative, positive or $0$. If you have (2) people playing tug of war the losing team performed negative work while the winning team performed positive work along the axis of the rope. If those people didn't change their elevation relative to the center of earth's mass, the gravity exerted by the center of mass for performed $0$ work while they were playing, even if they temporarily were dug into the ground (having a net displacement of $0$ results in a net work of $0$ - even if you temporarily had some displacement).
You could probably look at work as "when push came to shove how effective was the force at displacing"?
Assuming a stationary medium relative to an external observer, if you're sliding (relative to the stationary medium/observer) along some +x axis then any friction you experience would be accelerating you in the opposite direction, then the force friction delivers would be along the -x axis. Friction would only act on something moving to accelerate it in the opposite direction its current velocity (which is a vector, because velocity is speed and direction), so friction, from its perspective, always has a negative displacement it works over, so the work friction does is always negative.
If something is already moving really fast it's also possible that all forces perform negative work (at in a simplified model where you have a direction such that no force acts positively in it). Force has a say in the change in momentum once it starts acting on something, but doesn't have any say over what the momentum was prior to it acting on it, and for that reason the work done by that force can be $0$, negative or positive (as oppose to the net change in momentum which is in the same direction as the net force). Of course if you started with $0$ momentum, then the work done by a positive net force cannot be negative, and if there was any change in momentum the work by that net force will be positive.
Since work done by a force $\vec F$ undergoing a displacement $d\vec r$ is defined as $\vec F \cdot d\vec r$ when this dot product is positive the force and displacement are in the same direction and is negative when they are in opposite directions.
The work done by a frictional force does not always have to be negative.
Imagine a block $A$ on top of block $B$ and a force is applied to block $B$ to make both blocks increase their speed in a horizontal direction.
The frictional force on block $B$ due to block $A$ certainly does negative work because force is in the opposite direction to the displacement of block $B$.
However the frictional force on block $A$ due to block $B$ does positive work on block $A$ (increasing its kinetic energy) because the frictional force and displacement are in the same direction.
So decide on the direction of the force and the direction of its displacement and the definition of work done will do the rest.
You pull a spring to extend it. The force you exert on the spring is in the same direction as your displacement. You have done positive work on the spring.
The force that the spring exerts on you is in the opposite direction to its displacement so the spring has done negative work.
In physics when a force is applied on a body, work is done by the force on the body. It is a dot product of force and displacement it causes (say $d$). Then $$ W = F\cdot d = F d \cos (\theta) $$ where $\theta$ is the angle between the force and displacement vectors.
If the force is acting in one direction but displacement is in the opposite direction, the angle between these 2 vectors would be 180 degrees and $\cos(180^\circ)=-1$. Hence $$ W = F d \cos (180^\circ) = -Fd $$
You see that the work done is negative. However if the force and displacement were both in the same direction, the angle would have been zero and $\cos (0^\circ) = +1$. Work done would have been $W= Fd$ or a positive value. Hence positive work would have been done by the force.
A more tangible way of comprehending this topic (that can be a little confusing) is to imagine force acting in one direction but displacement happening in opposite direction. Such a situation can arise when a body is moving in a certain direction or the displacement is happening in a certain direction but a force comes and acts in the opposite direction. The force, you can imagine will reduce the velocity of the body. If this happens, the kinetic energy (KE) of the body will reduce. Since there is a reduction in KE, we say force has done negative work.
Alternately, if force and displacement are in the same direction, the force must be increasing the velocity of the body. Therefore the KE must be increasing. This increase in KE is associated with positive work done by the Force
Watch this video made by me for more clarity
When work done is in direction of force $\implies \theta<90$ is positive and work done is negative when $θ>90$ . In case of friction , you states that internal energy is increased. That internal energy is not necessary to be kinetic energy. Body gains internal energy by heat!
$\theta<90 \implies \cos\theta$ is positive. $90<\theta<180\implies \cos\theta$ is negative.
It is interesting to know while ( applying brakes ) to stop a moving vehicle , work done by friction is positive.
