Let's imagine there's a block of mass m sitting on a rough surface with a kinetic friction coefficient u. It's being pulled with a constant force A at h degrees above the horizontal and is displaced a distance d. The work done by A on the block is positive and is the horizontal component of A times the displacement:

$A\cos (h)*d$

I thought this meant that the work done by friction would be either the negative of that amount or else

$\mu*(mg-Asin (h))*d$

but apparently not?

Let's imagine there's a block of mass m sitting on a rough surface with a kinetic friction coefficient $\mu$. It's being pulled with a constant force $A$ at $h$ degrees above the horizontal and is displaced a distance $d$. The work done by $A$ on the block is positive and is the horizontal component of $A$ times the displacement:

$$A\cos (h)*d.$$

I thought this meant that the work done by friction would be either the negative of that amount or else

$$\mu*(mg-A\sin (h))*d$$

but apparently not?

Let's imagine there's a block of mass m sitting on a rough surface with a kinetic friction coefficient u. It's being pulled with a constant force A at h degrees above the horizontal and is displaced a distance d. The work done by A on the block is positive and is the horizontal component of A times the displacement:

A*cos(h)*d

I thought this meant that the work done by friction would be either the negative of that amount or else

u*(mg-Asin (h))*d

but apparently not?

Let's imagine there's a block of mass m sitting on a rough surface with a kinetic friction coefficient u. It's being pulled with a constant force A at h degrees above the horizontal and is displaced a distance d. The work done by A on the block is positive and is the horizontal component of A times the displacement:

$A\cos (h)*d$

I thought this meant that the work done by friction would be either the negative of that amount or else

$\mu*(mg-Asin (h))*d$

but apparently not?