Question

A 18.3 kg block is dragged over a rough, hor- izontal surface by a constant force of 114 N acting at an angle of 30◦ above the horizon- tal. The block is displaced 63.3 m, and the coefficient of kinetic friction is 0.134.
The acceleration of gravity is 9.8 m/s2

Find the work done by the 114 N force. Answer in units of J.
016 (part 2 of 3) 10.0 points
Find the work done by the force of friction.
Answer in units of J.
017 (part 3 of 3) 10.0 points
If the block was originally at rest, determine its final speed.
Answer in units of m/s.

Answer 1

someone please help me

Answer 2

Work is
\[W=Fd\]
Since the force is acting at an angle above the horizon at 30 deg
It will be\[F_{x}=F \cos (\theta)\]Since you know the distance this force act you can calculate the work done.
Friction is the same and you calculate that with f=cmg, where c is the coefficient of kinetic friction. Work done by friction\[Wf=cmgd\]
To calculate the velocity, you take the work done by the force W -Wf and that will give you the net energy left after friction was acting.
Since the block is moving it has kinetic energy which means you can use kinetic energy formula and get the velocity.\[W - Wf = \frac{ 1 }{ 2 }m v^{2}\]solve for v and you will have\[\sqrt{ \frac{ 2(W - Wf) }{ m }} =v\] From here you just plug in your numbers.