Work Done and Direction with frictional forces

Question

This might be a dumb question, but if work done is equal to the force exerted on an object in the direction of motion (so W = F*S), then how come frictional forces do work on an object. For example, in the picture below, if the object is moving on a rough horizontal plane and friction is acting on the object in the completely opposite direction to motion, then how is work being done (especially since negative work cannot be done as work is a scalar quantity).

Answer 1

Definition of power and work. The definition of the power of a force $\mathbf{F}$ ancting on a point moving with velocity $\mathbf{v}$ is

$$P := \mathbf{F} \cdot \mathbf{v} \ .$$

It's work while the system evolves from point $\mathbf{r}_0$ to point $\mathbf{r}_1$ along the path $\gamma$ is defined as $$W = \int_{\gamma} \mathbf{F} \cdot d\mathbf{r} = \int_{\gamma} \mathbf{F} \cdot \mathbf{v} \, dt = \int_{\gamma} P \, dt \ ,$$

having written the elementary displacement as the product of the velocity and the elementary time interval, $d \mathbf{r} = \mathbf{v} d t$.

Remark: power of a force can be negative. Being the scalar product of two vector quantities, power is negative if the angle between these two vectors is larger than $\pi/2$ (and in 2D problems $\theta \in [\frac{\pi}{2}, \frac{3}{2}\pi]$), so that $\cos \theta < 0$. Work can be negative has well, being the integral of a function that can be negative.

Power of dynamic friction. The total power of dynamic friction exchanged between two parts of a system is always negative, and it dissipates mechanical energy into heat. Being the power always negative, the work is negative as well. As suggested by Dale in his comment, if friction acts as an external force on the system, it can have add positive power to the system. See the end of the answer for a discussion of two bodies in relative motion.

Focusing on the motion of a body sliding on a wall at rest, the dynamic friction force is always opposite w.r.t. the velocity of the body. With some math, if the velocity of the body is $\mathbf{v} = v \mathbf{\hat{x}}$, the dynamic friction can be modeled as $\mathbf{F}^d = - f^d N \frac{\mathbf{v}}{|\mathbf{v}|} = -f^d N \, \text{sgn}(v)\mathbf{\hat{x}}$, with $f^d$ dynamic friction coefficient and $N\ge0$ normal reaction from the wall.

From the definition of the power of a force, $P = \mathbf{F} \cdot \mathbf{v}$, the power of the friction force reads

$$P^d = \mathbf{F}^d \cdot \mathbf{v} = - f^d N \frac{\mathbf{v}}{|\mathbf{v}|} \cdot \mathbf{v} = - f^d N |\mathbf{v}| \le 0 \ .$$

Edit, following Dale's comment. Friction between two bodies in relative motion. Let's consider here an example of two bodies in relative motion w.r.t. an inertial observer, sliding one over the other on an interface with dynamical friction coefficient $f^d$. Let body $1$ and body $2$ move with velocity $\mathbf{v}_1 = v_1 \mathbf{\hat{x}}$ and $\mathbf{v}_2$ w.r.t. the inertial observer. Let $N$ the norm of the normal component of the force exchanged and $\mathbf{F}^d_{12}$ the dynamical friction acting on body $1$ due to the friction with body $2$, and $\mathbf{F}^d_{21}$ the force acting on $2$ due to $1$.

Force $\mathbf{F}^d_{12}$ is opposite w.r.t. the relative velocity $\mathbf{v}_{1/2} := \mathbf{v}_1 - \mathbf{v}_2 = - (\mathbf{v}_2 - \mathbf{v}_1) = - \mathbf{v}_{2/1}$,

$$\mathbf{F}^d_{12} = - f^d N \frac{\mathbf{v}_{1/2}}{|\mathbf{v}_{1/2}|} = - \mathbf{F}^{d}_{21} \ .$$

While the power of friction acting on the individual bodies could be either positive or negative,

$$P^{d}_1 = \mathbf{F}^d_{12} \cdot \mathbf{v}_1 = - f^d N \frac{\mathbf{v}_{1/2}}{|\mathbf{v}_{1/2}|} \cdot \mathbf{v}_1$$ $$P^{d}_2 = \mathbf{F}^d_{21} \cdot \mathbf{v}_2 = f^d N \frac{\mathbf{v}_{1/2}}{|\mathbf{v}_{1/2}|} \cdot \mathbf{v}_2 \ ,$$

the total power $P^d = P^d_1 + P^d_2$ is always negative,

$$P^d = - f^d N \frac{\mathbf{v}_{1/2}}{|\mathbf{v}_{1/2}|} \cdot (\mathbf{v}_1 - \mathbf{v}_2) = - f^d N |\mathbf{v}_{1/2}| < 0 \ .$$