Question

An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ft-lbf and an increase in potential energy of 1500 ft-lbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are 40 ft/s and 30 ft, respectively.

Required:
a. Find final velocity in ft/s.
b. Find final elevation.

Answer 1

a) the final velocity is 35.75 ft/s

b) The final elevation is 45 ft

Step-by-step explanation:

Given the data in the question;

Weight of object; W = 100 lbf

Change in kinetic energy; ΔE = 500 ft-lb

so

`(1)/(2)`m
`v_i^2` -
`(1)/(2)`m
`v_f^2` = ΔE

`(1)/(2)`m
`v_i^2` -
`(1)/(2)`m
`v_f^2` = 500

multiply both sides by 2

m
`v_i^2` - m
`v_f^2` = 1000

m(
`v_i^2` -
`v_f^2` ) = 1000

`v_i^2` -
`v_f^2` = 1000/m

`v_i^2` -
`v_f^2` = (1000)(g) / W

we know that, acceleration due to gravity g = 9.8 m/s² = 32.18 ft/s²

so we substitute

`v_i^2` -
`v_f^2` = (1000)(32.18) / 100

`v_i^2` -
`v_f^2` = (1000)(32.18) / 100

`v_i^2` -
`v_f^2` = 32180 / 100

`v_i^2` -
`v_f^2` = 321.8

since The initial velocity
`v_i` is given to be 40 ft/s;

(40)² -
`v_f^2` = 321.8

1600 -
`v_f^2` = 321.8

`v_f^2` = 1600 - 321.8

`v_f^2` = 1278.2

`v_f` = √1278.2

`v_f` = 35.75 ft/s

Therefore, the final velocity is 35.75 ft/s

b)

we know that;

change in potential energy is;

ΔP.E = mg( h
`_f` - h
`_i` )

given that; increase in potential energy; ΔP.E = 1500 ft-lbf

and mg = Weight = 100 lbf

we substitute

1500 = 100( h
`_f` - h
`_i` )

h
`_f` - h
`_i` = 1500 / 100

h
`_f` - h
`_i` = 15 ft

given that, elevation of the object; h
`_i` = 30 ft

h
`_f` - 30 ft = 15 ft

h
`_f` = 15 ft + 30 ft

h
`_f` = 45 ft

Therefore, The final elevation is 45 ft