Work energy and power M2

It's the same as in the first two parts, except the directions are reversed, so tack on a negative sign or resolve in the opposite direction (horizontally), preferably the latter. Your reaction force thingy will also change because the force is pushing in the same direction (vertically-wise) as the weight.

Show us your working.

When you're given a force pointing inwards into an object, then just extend that line to it points outward from the other side of the object to make it easier. Then resolve vertically and horizantally for both R and the Net Force, use F=ma to find a, then use s = ut + .5at^2 to find s. Then just use Wd = F*d.

What was your answer? I'll cross check.

When you're given a force pointing inwards into an object, then just extend that line to it points outward from the other side of the object to make it easier.

There is a vertical component, in this case, if you resolve downwards, you get: $100 \sin 20^{\circ} + mg - R = 0$

Noooo! mg=R is not true. In either case- the vertical component is useless here, the box can only move sideways, use only horizontal components of force.

Done. Check out my working.

Ah shizzle, i just read the question properly. You need to find the reaction force R which is mg + 100 sin 20. And then uR is the frictional force.

So your resultant horizontal force is 45 (don't round off btw) - uR.

Done. Check out my working.