A block of mass 1.00 kg sits on an inclined plane as shown. (Figure 1) A force o
ID: 1396358 • Letter: A
Question
A block of mass 1.00 kg sits on an inclined plane as shown. (Figure 1) A force of magnitude 100 N is pulling the block up the incline. The coefficient of kinetic friction between the plane and the block is 0.500. The inclined plane makes an angle 40.0 degrees with the horizontal.
Part A
What is the total work Wfric done on the block by the force of friction as the block moves a distance 9.00 m up the incline?
Express the work done by friction numerically, in Joules.
Part B
What is the total work WF done on the block by the applied force F? as the block moves a distance 9.00 m up the incline?
Express your answer numerically, in Joules.
Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline. (Figure 2)
Part C
What is the total work Wfric done on the block by the force of friction as the block moves a distance 9.00 m down the incline?
Express your answer numerically, in Joules.
Part D
What is the total work WF done on the box by the applied force 100 N in this case?
Express your answer numerically, in Joules.
Explanation / Answer
Here ,
mass of block , m = 1 kg
theta = 40 degree
coefficient of kinetic friction , u = 0.500
A)
work done by friction , Wfric = - friction force* distance
work done by friction , Wfric = - u*mg* cos(theta) * distance
work done by friction , Wfric = - 1 * 0.5 * 9.8 * cos(40) * 9
work done by friction , Wfric = -33.8 J
the work done by friction , Wfric is -33.8 J
B)
Work done by applied force , WF = F*d
Work done by applied force , WF = 100 * 9
Work done by applied force , WF = 900 J
the Work done by applied force , WF is 900 J
C)
Now, as the friction will act in oposite direction to motion
work done by friction , Wfric = - friction force* distance
work done by friction , Wfric = - u*mg* cos(theta) * distance
work done by friction , Wfric = - 1 * 0.5 * 9.8 * cos(40) * 9
work done by friction , Wfric = -33.8 J
the work done by friction , Wfric is -33.8 J
D)
as displacement and acting force are in same direction
Work done by applied force , WF = F*d
Work done by applied force , WF = 100 * 9
Work done by applied force , WF = 900 J
the Work done by applied force , WF is 900 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.