ka A 4.80-kg block is set into motion up an inclined plane with an initial speed

Question

ka

A 4.80-kg block is set into motion up an inclined plane with an initial speed of vi 8.40 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 e 30.0% to the horizontal. (a) For this motion, determine the change in the block's kinetic energy. Enter a number. change in a quantity is defined as the final value minus the initial value. J (b) For this motion, determine the change in potential energy of the bloc system. k-Earth (c) Determine the friction force exerted on the block (assumed to be constant) (d) What is the coefficient of kinetic friction?

Explanation / Answer

a)

change in energy = 0.5*4.8*8.4^2 = -169.34 J

b)

change = 4.8*9.8*3*sin(30 degrees) = 70.56 J

c)

friction = (169.34-70.56)/3 = 32.9 N

d)

coefficient = 32.9/(4.8*9.8*cos(30 degree)) = 0.807

Related Questions

Navigate

• Browse (All)
• Browse K
• Subjects

---

---

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