At t=0 a 1.0kg ball is thrown from a tall tower with v=(18 m/s)i+(24 m/s)j. what is ΔU of the ball-earth system between t=0 and t=6.0s (still free fall)Δ

Respuesta :

Answer:

ΔU = 3492.72 J

Explanation:

Given

m = 1.0 kg

v = 18 i m/s + 24 j m/s

To find the initial distance to determine the initial potential energy can use newton's equations

[tex]v = \sqrt{18^2 + 24 ^2} = 30m/s[/tex]

[tex]s =  v_o*t + \frac{1}{2}*a *t^2[/tex]

[tex]s = 30 m/s * 6 s + 4.9 m/s^2 * 36 s^2 = 356.4 m[/tex]

Now the potential energy can be find using

[tex]E_p = m*g*h[/tex] and final energy as t=6 [tex]E_p = m*g*(h-356.4m)[/tex]

Initial and final energy can be related as:

ΔU = m*g*h - m*g*(h-356.4m)

So get the factor to solve the total energy

ΔU = m*g * (h - h + 356.4)

ΔU = 1.0 kg * 9.8  m/s^2 * 356.4 m

ΔU = 3492.72 J