A block is given an initial velocity up an inclined plane with friction. The block comes to rest before the end of the inclined plane.
What is the sign of the work done by gravity as the block comes to rest on the inclined plane?

The sign of the work done by gravity on a block moving up an inclined plane as the block comes to a rest on the inclined plane is **negative**.

Explanation:

Work is given mathematically as (Force).(displacement) = /F/ /d/ cos θ

where /F/ = magnitude of the force

/d/ = d = magnitude of displacement that the force moves through

θ = angle between the force and the displacement

For gravity Force,

Whenever the angle between the force and the displacement is in the range of 0° and 90°, the workdone is positive.

And whenever the angle between the force and the displacement is in the range of 90°+ to 180°. The workdone is negative.

Better explained,

For gravity, the force of gravity acts in the negative y direction, so, force of gravity is always equal to (-mg î).

If an object is falling downwards, then its displacement is in the negative y direction too; - dî.

Work done by gravity on a falling object = (-mgî).(-dî) = + mg d = mgd cos 0° = + mgd (θ = 0°) in this case

Positive work!

And for an object rising upwards, the force of gravity is still in the negative y direction too and is equal to (-mgî). But the displacement is in the positive y direction; that is, +dî

Work done by gravity on a body moving upwards = (-mgî).(dî) = - mgd = mgd cos 180° = - mgd (θ = 180°)

Negative work done!

So, for a body moving up an inclined plane, the vertical displacement is still upwards and in the positive y-direction. So, the analogy of the 2nd gravity explanation works for it.

0.11 m/s and 0.28 m/s should be correct