A rock with mass m = 2.10 kg falls from rest in a viscous medium. The rock is ac

Question

A rock with mass m = 2.10 kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of F = 15.1 N (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force f=kv, where v is the speed in m/s and k = 2.20 N×s/m .

1)Find the coordinate 1.50 s after the start of the motion

2)Find the speed 1.50 s after the start of the motion

3)Find the acceleration 1.50 s after the start of the motion.

4)Find the time required to reach a speed 0.9vt

Explanation / Answer

1.)
Coordinate 1.50 s after the start of the motion -
S = ut + 0.5 at^2
Initial Velcoity, u = 0 9falls from rest)
S = 0.5*at^2 = 0.5*a*1.5^2
S = 1.125 a
S =

vy(1.5s) = vt[1-e^(-kt/m)]

Where vt is terminal speed
15.1 = Vt * 2.20
Vt = 6.86 m/s

vy(1.5s) = 6.86m/s * [1-e^(-2.2*1.5/2.1)]
vy(1.5s) = 5.43 m/s

y(t) = integral  vt[1-e^(-kt/m)]
y(t) = vt * (t + 0.95 * e^(-1.05t))
y(1.5) = 6.86 * (1.5 + 0.95 * e^(-1.05*1.5))
y(1.5) = 11.64 m

ay(t) = dvy/dt
ay(t) = d/dt(vt[1-e^(-kt/m)])
ay(t) =  d/dt(6.86 * [1-e^(-1.05t)])
ay(t) = 6.86 * 1.05 * e^-1.05t
ay(1.5) = 6.86 * 1.05 * e^(-1.05*1.5)
ay(1.5) = 1.49 m/s^2

0.9 * 6.86 = 6.86* (1 - e^(-1.05t))  
0.9 = (1 - e^(-1.05t))  
solving for t
t = 2.2 s

1.) y(1.5) = 11.64 m
2.) vy(1.5s) = 5.43 m/s
3.) ay(1.5) = 1.49 m/s^2
4.) t = 2.2 s

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