A package is pushed up an incline at x = 0 with an initial speed v 0 . The incli

Question

A package is pushed up an incline at x = 0 with an initial speed v0. The incline is coated with a thin viscous layer so that the acceleration of the package is given by a = -(g sin ? + Nv), where g is the acceleration due to gravity, N is a constant, and v is the velocity of the package.

If ? = 25

A package is pushed up an incline at x = 0 with an initial speed v0. The incline is coated with a thin viscous layer so that the acceleration of the package is given by a = -(g sin ? + Nv), where g is the acceleration due to gravity, N is a constant, and v is the velocity of the package. If ? = 25degree, v0 = 1.8288 m/s, and N = 5 s^-1, determine the distance d traveled by the package before it comes to a stop.

Explanation / Answer

dv/dt = a = -(gsin(theta)+nv)

then integrating w.r.t t, then dv/(gsin(25)+5v) = dt, then ln(4.15+5v) = -t + k k is integration constant

at t=0, v=1.8288m/s, then k = 2.59, then ln(4.15 + 5v) = -t + 2.59

v = -..83 + 2.6588e^-t = dx/dt

again if we integrate, x = -.83t - 2.6588e^-t + m where m is constant

at t=0, x=0, thenn m = 2.6588

then x = -.83t - 2.6588*e^-t + 2.6588

at t=1.1642 sec, v = 0, then x = .8625 m

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