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3 years ago

A 0.5 kg ball moving at a speed of 3m/s rolls up a hill. How high does the ball roll before it stops

1 answer:

3 0

According to conservation of energy
Kinetic energy at the bottom of hill = potential energy at top of hill
Kinetic energy(KE)= 0.5×mass(m)×velocity(v)²
where m=0.5 kg, and v= 3 m/s, v²=9
KE=0.5×0.5×9=2.25 J = PE
PE=m×gravitational acceleration(g)×height(h)
h=2.25/(0.5×9.8)=0.459 m

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