A 12.5kg crate slides along a horizontal frictionless surface at a constant spee

Question

A 12.5kg crate slides along a horizontal frictionless surface at a constant speed of 4.0m/s. The crate then slides down a frictionless incline and across a second horizontal surface. A: what is the kinetic energy of the crate as it slides on the upper surface? B: while the crate slides along the upper surface, how much gravitational potential energy does it have compared to what it would have on the lower surface? C: what is the speed of the crate when it arrives at the lower surface? E: what minimum coefficient of kinetic friction is required to bring the crate to a stop over a distance of 5.0m along the lower surface?

Explanation / Answer

KE= 1/2 mv^2 = 0.5 X 12.5 X 4 X4= 100 J a. 100J b. it has a PE = mgh c. speed will be greater, = v+ sqrt(2gh) d. 100+ mgh = nmg X 5 n= co eff of friction. values of h is not given so, put any value of h required in the above relations

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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