NOTE: This question has already been posted, but it looks like the answer provid

Question

NOTE: This question has already been posted, but it looks like the answer provided was incorrect.

Am I correct in thinking A. is 0 Nm, because the Net force = 0 N (since there's a constant velocity therefore a = 0 m/s^2)?

30 pts.] A block is on a plane inclined at an angle a-30° from the horizontal. The coeffcint f betweenthe block and the incline is .-0.12. A force is applied along the incl net ablock of ass mn2kgass own in the figure, and the mass moves up the incline at constant velocity c friction v2 m/s. (SHOW YOUR WORK!) (a) What is the net work done on the box over a displacement of 1.2m? What is the work done by the force F over this displacement? (c) What is the magnitude of the force F? =30

Explanation / Answer

(a) Since the box is moving with constant velocity, means it has no acceleration.

So, F = mg*sin30 + u*mg*cos30 = 2*9.81*0.5 + 0.12*2*9.81*cos30 = 9.81 + 2.04 = 11.85 N

So, the net work done on the box over a displacement of 1.2 m = F*d = 11.85*1.2 = 14.22 Joule

(b) Work done by the force is the same as calculated above = 14.22 J

(c) Magnitude of force F = 11.85 N (as calculated above).

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.