A 4.80-kg particle moves along the x axis. Its position varies with time accordi

Question

A 4.80-kg particle moves along the x axis. Its position varies with time according tox = t + 5.0t, where x is in meters and t is in seconds. (a) Find the kinetic energy of the particle at any time t. (Use the following as necessary: t.) (b) Find the magnitude of the acceleration of the particle and the force acting on it at time t. (Use the following as necessary: t) (c) Find the power being delivered to the particle at time t. (Use the following as necessary: t.) (d) Find the work done on the particle in the interval t0 to 3.10 s. Need Help? ?inen Intern Show My Work (optional) O -1 points SerPSET9 8.P.063.MI. My Notes A 10.0-kg block is released from rest at point in the figure below. The track is frictionless except for the portion between points B and , which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2 350 N/m, and compresses the spring 0.200 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points and C. 3.00 m 6.00 m

Explanation / Answer

Given that x = t+ 5t^3

v = dx/dt = 1+(15*t^2)

a) K = 0.5*m*v^2 = 0.5*4.8*(1+15t^2)^2

b) a = dv/dt = 30*t

F = m*a = 4.8*30*t = 144*t

C) P = F*V = 144*t*(1+(15t^2)) = 144t + 2160*t^2

d) WORK DONE IS W = change in KInetic energy = Kf-Ki

W = 0.5*m*(v^2-u^2)

v = 1+(15*t^2) = 1+(15*3.1^2) = 145.15 m/s

u = 1+(15*0^2) = 1 m/s

W = 0.5*4.8*(145.15^2-1^2) = 50562 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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