A worker pushes horizontally on a large crate with a force of 170 N, and the crate is moved 3.6 m. How much work was done
1 answer:
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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= 0.5×200×0=0 j
PEi= mgh = 200×9.8×0=0 j
KEi= 1/2 m=0.5×200×
=100
j
Put these values in eq 1, we get;
0-100 +3018.4-0=0
-100 =-3018.4
==> = 30.184
==> Vi =