Question
A block of ice of mass 30 kg is at rest atop an inclined plane of vertical height
2 m. It is released and slides down the inclined plane. What is its speed at the bottom of the inclined plane?
2 m. It is released and slides down the inclined plane. What is its speed at the bottom of the inclined plane?
Answers
speedbottom:
KE at bottom=PE at top
1/2 mv^2=mgh
v^2=2gh=2*9.8*2
speedbottom=sqrt(4*9.8) m/s
KE at bottom=PE at top
1/2 mv^2=mgh
v^2=2gh=2*9.8*2
speedbottom=sqrt(4*9.8) m/s
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