Answer:
(a). Energy is 64,680 J
(b) velocity is 51.43m/s
(c) velocity in mph is 115.0mph
Explanation:
(a).
The potential energy [tex]P[/tex] of the payload of mass [tex]m[/tex] is at a vertical distance [tex]h[/tex] is
[tex]P =mgh[/tex].
Therefore, for the payload of mass [tex]m = 50kg[/tex] at a vertical distance of [tex]h = 132 m[/tex], the potential energy is
[tex]P = (50kg)(9.8m/s^2)(132m)[/tex]
[tex]\boxed{P = 64,680J}[/tex]
(b).
When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,
[tex]mgh= \dfrac{1}{2}mv^2[/tex]
[tex]v= \sqrt{2gh}[/tex]
[tex]v = \sqrt{2*9.8*135}[/tex]
[tex]\boxed{v = 51.43m/s}[/tex]
(c).
The velocity in mph is
[tex]\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}[/tex]
[tex]\boxed{v= 115.0mph}[/tex]