Answer:
5.5 m/ sec
Explanation:
Because the inclined surface is frictionless so we can assume that total change of energy is zero
i-e ΔE = 0
Or we can say that difference between final and initial energy is zero i-e
Ef- Ei =0
Where,
Ef= final energy at the top of the ramp= KEf+PEf
Ei= Initial energy at the bottom of the ramp=KEi+PEi
So we have
(KEf+PEf)-(KEi+PEi)=0
==>KEf-KEi+PEf-PEi=0 -------------(1)
KEf = mgh = 200×9.8×h
Where h= Sin 22 = h/d= h/4.1
or
0.375×4.1=h
or h= 1.54 m
So, PEf= 200×9.8×1.54=3018.4 j
and KEf= 1/2 m[tex]Vf^{2}[/tex]= 0.5×200×0=0 j
PEi= mgh = 200×9.8×0=0 j
KEi= 1/2 m[tex]Vi^{2}[/tex]=0.5×200×[tex]Vi^{2}[/tex]=100[tex]Vi^{2}[/tex] j
Put these values in eq 1, we get;
0-100 [tex]Vi^{2}[/tex]+3018.4-0=0
-100 [tex]Vi^{2}[/tex]=-3018.4
==> [tex]Vi^{2}= \frac{3018.4}{100}[/tex] = 30.184
==> Vi = [tex]\sqrt{30.184} = 5.5 m.sec[/tex]
