Physics Review Question

Question

The figure below shows a block of mass m=2.0 sliding to the right across a floor through a distance of 20.0 cm the coefficient of kinetic friction between the floor and block 0.15

Had to get creative because I couldn't attach picture.

---------------------------

\ (Theta)

F \

\

(Block) (Force F is attached to top left of block)

-------------------------------- (Ground)

The applied force F has a magnitude 4.00 and is directed at 60 degrees below the horizontal

What is the work done on the block by :

Gravitational

Normal Force

The Force F

Force of Friction on Block

Answer 1

Well, first of all, good job explaining the image above. There are a few issues, though. Its not clear if the force on top left is pushing the block, or pulling it. Given that the thing at top left looks like a string, it could be a pull. But since the block is moving to the right side and I dont see any other force, I'll assume its a push.

Right off the bat, since the height of the block isn't changing, the work done by gravity as well as normal force is zero. Now the force applied by the at the top left has the horizontal component of F sin(theta) and vertical component of F*cos(theta). Calculate the total force on the block downward:

F_down= m*g + F* cos (theta)

Your work by friction force will be = μ*F_down*displacement

And the work done by the force F only needs the horizontal component, since thats the direction of motion. So

Work by force F = F sin(theta) * displacement.

Substitute the values from the question to get your answers!

If you need help understanding any of the concepts I used above, please feel free to message me on my profile here.