A 5.80-kg block is set into motion up an inclined plane with an initial speed of

Question

A 5.80-kg block is set into motion up an inclined plane with an initial speed of vi = 7.80 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of theta = 30.0 degree to the horizontal. For this motion, determine the change in the block's kinetic energy. J For this motion, determine the change in potential energy of the block-Earth system. J Determine the friction force exerted on the block (assumed to be constant). N What is the coefficient of kinetic friction?

Explanation / Answer

change in potential energy =m*g*(dsin)=5.8*9.8*3sin30=85.26

change in kinetic energy= 0.5*m*(v^2-0^2)=176.436

friction force=176.436-85.26=91.176

s=91.176/(5.8*9.8cos30)=1.8522339

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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