To push a 4-kg crate up a frictionless incline angled at 15° to the horizontal,
ID: 1551839 • Letter: T
Question
To push a 4-kg crate up a frictionless incline angled at 15° to the horizontal, a worker exerts a force of 200 N parallel to the incline. The crate moves a distance of 6 m.
What work is done on the crate by the worker?
What work is done on the crate by gravity?
What is the work done on the crate due to the normal force of the inclined plane surface?
What is the net work done on the crate?
If the crate begins at rest, what is the final speed of the crate as it is moved from the bottom to the top of the inclined plane?
Explanation / Answer
A.
Mechanical work is Force*distance
W=200*6=1200J
B.
The work of the gravity is the change in potential energy
W=-m*g*h=4*9.81*6*sin(15)=-60.93J (movement is opposite to the direction of the gravitation!)
C.
Obviously 0. The vector is perpendicular to the direction of movement.
D.
Energy conserves.
The net work of all forces will be input work of the worker + work of gravitation.
W=1200+(-60.93)= 1139.07 J
That amount of energy is obviously the change in kinetic energy (crate is accelerated as it moves).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.