SOLUTION (A) Find the work done by friction on the sled and the net work, if the
ID: 1345425 • Letter: S
Question
SOLUTION
(A) Find the work done by friction on the sled and the net work, if the applied force is horizontal.
First, find the normal force from they-component of Newton's second law, which involves only the normal force and the force of gravity.
Fy = n ? mg = 0 ? n = mg
Use the normal force to compute the work done by friction.
Sum the frictional work with the work done by the applied force to get the net work (the normal and gravity forces are perpendicular to the displacement, so they don't contribute).
(B) Recalculate the frictional work and net work if the applied force is exerted at a 30.0° angle.
Find the normal force from the y-component of Newton's second law.
Use the normal force to calculate the work done by friction.
Sum this answer with the work done by the applied force to get the net work (again, the normal and gravity forces don't contribute).
LEARN MORE
REMARKS The most important thing to notice here is that exerting the applied force at different angles can dramatically affect the work done on the sled. Pulling at the optimal angle (11.3° in this case) will result in the most net work for the same applied force.
QUESTION By what factor is the net work multiplied in each case if the displacement is doubled?
PRACTICE IT
(b) Repeat the calculation if the applied force is exerted at an angle of ? = 30.0° with the horizontal.
Explanation / Answer
Solution:
PRACTICE IT
a)
Work done by friction is ,
Wf = - uk mg*S = - 0.2*50*9.81*5.10 = - 500.31J
Total work done,
Wnet = WF + Wf = [125*5.10] - 500.31 = 137.19J.
b)
Work done by friction remians contant.
Wf = -500.31J
Total Work done,
Wnet = WF + Wf = [125*5.10*Cos30o] - 500.31
= 51.78 J.
I hope you understood the problem, If yes rate me!! or else comment for a better solution.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.