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

Answer:

a) the final velocity is 35.75 ft/s

b) The final elevation is 45 ft

Explanation:

Given the data in the question;

Weight of object; W = 100 lbf

Change in kinetic energy; ΔE = 500 ft-lb

so

$\frac{1}{2}$m$v_i^2$ - $\frac{1}{2}$m$v_f^2$ = ΔE

$\frac{1}{2}$m$v_i^2$ - $\frac{1}{2}$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