Question

A block is given n intial speed of 3.0 s^-1 m up the 22.0 degree plane

(a) How far up the plane will it go?

(b) How mcuh time elapses before it returns to its starting point?

Assume Mu k = 0.17

for (a)

a = - (Mu k g cos theta + g sin theta)

I got - 5.23 s^-1 m

for finding distance

x = (2a)^-1 (V^2 - Vo^2)

I got .87 m

ok in my book it gives me the answers in centimeters but still got right 87 cm for the time it gives me 1.5 s don't know how to get the time

t = a^-1 (V - Vo)
multiplied by two for time back
t = 1.15 s

don't know what to do

thanks

Answer 1

The average speed going up is 3/2 or 1.5m/s, so time up is .87/1.5 sec

Now going down, it has .87*sin22 height, so it willhave some vfinal at the bottom

1/2 m vf^2=mg (.87sin22)-.87*mgCos22
which means final KE=initial PE-friction

solve for vf, then avg velocity down is 1/2 vf, and then solve for time down, and add to time up.