Respuesta :

paulandigi

Solve the

potential energy at point B


PE = mgh


Where m is

the mass


G is the

acceleration due to gravity 9.8 m/s2


H is the height


PE = ( 1.5

kg) ( 9.8 m/s2) (0.5 m)


PE = 7.35 J



Solve the

velocity using the


KE = 0.5mv^2


Where ke is

the kinetic energy


M is the

mass


V is the

velocity


Since all

energy is converted to KE


So KE = PE



7.35 = 0.53(1.5)

v^2


V = 3.13 m/s


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Explanation:

It is given that,

Mass of the ball, m = 1.5 kg

The ball is pulled to a height of 0.5 meters and then released.

The potential energy at point B is given by :

[tex]E_p=m\times g\times h[/tex]

[tex]E_p=1.5\times 9.8\times 0.5[/tex]

[tex]E_p=7.35\ J[/tex]

To find the velocity at point A, it is required to use the conservation of energy as :

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\times 0.5}[/tex]

v = 3.13 m/s

Hence, this is the required solution.

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