The student pulls with a force of 200N a an angle of 45 degrees above the horizontal. The box mass of 50.0Kg and coefficient of kinetic friction between the box and the floor is 0.2. Find the acceleration of the box? Help.

Question

anonymous · Answer

Yes Jenny. Allow me to show you the sketch of forces first.

anonymous · Answer

Oops. How do I send a picture? Cannot find the tools in the left hand side of 'post'.

anonymous · Answer

Anyway, if you feel like needing the diagram, let me know. 
First thing you have to do is resolve horizontally in order to find the normal comtact force (R) which is above the horizontal axis. So above the horizontal axis or above the box, you have R and the 200N force acting in an angle of 45degrees to the vertical/horizontal. And below the box, you have the weight, which is equal to mg or 50*9.81=490.5N.
So solving vertically, you will have : R+200sin(45)=490.5, because some of forces above the box have to be equal to those below it. Your R will be 349.1N.
Secondly, you have to find the frictional force (Ff), because you are given the coefficient of friction. Given that Ff=muR, replacing values you will have: Ff=0.2*349.1=69.82N.
Finally, to find the acceleration, we use the very common forces formula to find the acceleration: F-Ff=ma, where F is the driving force in the horizontal direction, which the 200N one multiplied by cos(45), because you are resolving it horizontally.
Where Ff is the frictional force
a is the acceleration and 
m is the mass of the box.
200cos(45)-69.82=50a making a subject of the formula should give you 1.43m/s^2.
Hope you understood.