Question

A crate is given an initial speed of 3.1 m/s up the 29 degree plane shown in the figure . Assume coefficient of friction = 0.17.

How far up the plane will it go?

How much time elapses before it returns to its starting point?

Answer 1

KE= W(fr) +PE
KE =m•v²/2
W(fr) =F(fr) •s = μ•m•g•s •cosα
PE=m•g•h = m•g•s•sinα
m•v²/2 = μ•m•g •s •cosα + m•g•s•sinα =
=m•g•s(μ•cosα + sinα).
s= v²/2•g•(μ•cosα + sinα)=
=3.1²/2•9.8•(0.17•0.875+0.48)=0.78 m.
a1 = v²/2•s = 3.1²/2•0.78 =6.16 m/s².
0=v –a•t1
t1=v/a=3.1/6.16 = 0.5 s. (the time of upward motion)
m•a=m•g•sinα - μ•m•g • cosα
a2= g(sinα - μ•cosα)=9.8•(sin29-0.17•cos29)=3.25 m/s².
s=a2•t2²/2,
t2=sqrt(2s/a2) = sqrt(2•0.78/3.25)=0.69 s
t=0.5+0.69=1.19 s.