1.2.1Ep=mgh
=(2)(9,8)(12)
=235,2J
1.2.2Ep=Ek
Ek=235,2J
1.2.3 Ek=1/2mv^2
235,2 =(1/2)(2)v
(235,2)^1/2=(V^2)^1/2
15.336m/s=v
1.2.4 V^2=U^2 + 2as
20^2=0 + (2)(9.8)s
400=19,6s
400/19,6=s
20,408m=s
1.a ball with a mass 2kg rests on an incline with an angle of 10 degrees. The ball is released to roll down the incline plane neglecting all friction:
Determine the following:
A.the loss in potential energy after it has rolled 12m.
B. The kinetic energy
C.the velocity after it has rolled 12m.
D.the original height that the ball ,as rolled from in order to reach the bottom of the slope at 20m/s
3 answers
But tshepo that 12m is not a Height ita a Distance
You have to calculate height first
You have to calculate height first
What about if we say EP=mgh12sin10
=2*9.8*12sin10
=40,84J
=2*9.8*12sin10
=40,84J