Question

An object whose weight is 100lbf( pound force) experiences a decrease i kinetic energy of 500ft-lb, and an increase in potential energy of 1500ft-lbf. The initial velocity and elevation of the object, each relative to the surface of the Earth are 40ft/s and 30ft.
(a). Find final velocity in ft/s . (b) find final elevation.

Answer 1

Answer:

A. 43.84 ft/s.

B. 45 ft.

Explanation:

A.

Given:

Weight, W = 100 lbf

mass, M = 100/32.174

= 3.108 lb

kinetic energy, ΔKE = -500 ft.lbf

Potential energy ΔPE= 1500 ft.lbf

initial velocity, u  = 40 ft/s

initial height, h1 = 30 ft/s

acceleration, g = 32.174 ft/s^2

ΔKE = 1/2 * M * (v^2 - u^2)

-500 = 1/2 * 3.108 * (v^2 - 40^2)

v = 43.84 ft/s.

B.

ΔPE = M * g * (h2 - h1)

1500 = 3.108 * 32.174 * (h2 - 30)

h2 = 45 ft.

Answer 2

Answer:

a) 35.75 ft/s

b) 45 ft

Explanation:

Given  

Weight W = 100 lbf

mass(m) = 100*32.174/32.2=99.92 lb

decrease in kinetic energy ΔKE = -500 ft.lbf

increase in kinetic energy ΔPE= 1500 ft.lbf

initial velocity V_1  = 40 ft/s

initial height h_1 = 30 ft/s

The gravitational acceleration g = 32.2 ft/s2 Required  

(a) Final velocity V_2 (a) Final elevation h_2  

Solution

Change in kinetic energy is defined by

ΔKE = .5*m *( V_2 ^2-V_1^2)

Change in potential energy is defined by

ΔKE = W *( h_2 -h_1 )

Then,

-500=.5*99.92*1/32.174*(V_2 ^2-40^2)

V_2=35.75 ft/s

1500 = 100 x (h_2  — 30)  

h_2= 45 ft