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
[tex]\frac{1}{2}[/tex]m[tex]v_i^2[/tex] - [tex]\frac{1}{2}[/tex]m[tex]v_f^2[/tex] = ΔE
[tex]\frac{1}{2}[/tex]m[tex]v_i^2[/tex] - [tex]\frac{1}{2}[/tex]m[tex]v_f^2[/tex] = 500
multiply both sides by 2
m[tex]v_i^2[/tex] - m[tex]v_f^2[/tex] = 1000
m( [tex]v_i^2[/tex] - [tex]v_f^2[/tex] ) = 1000
[tex]v_i^2[/tex] - [tex]v_f^2[/tex] = 1000/m
[tex]v_i^2[/tex] - [tex]v_f^2[/tex] = (1000)(g) / W
we know that, acceleration due to gravity g = 9.8 m/s² = 32.18 ft/s²
so we substitute
[tex]v_i^2[/tex] - [tex]v_f^2[/tex] = (1000)(32.18) / 100
[tex]v_i^2[/tex] - [tex]v_f^2[/tex] = (1000)(32.18) / 100
[tex]v_i^2[/tex] - [tex]v_f^2[/tex] = 32180 / 100
[tex]v_i^2[/tex] - [tex]v_f^2[/tex] = 321.8
since The initial velocity [tex]v_i[/tex] is given to be 40 ft/s;
(40)² - [tex]v_f^2[/tex] = 321.8
1600 - [tex]v_f^2[/tex] = 321.8
[tex]v_f^2[/tex] = 1600 - 321.8
[tex]v_f^2[/tex] = 1278.2
[tex]v_f[/tex] = √1278.2
[tex]v_f[/tex] = 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[tex]_f[/tex] - h[tex]_i[/tex] )
given that; increase in potential energy; ΔP.E = 1500 ft-lbf
and mg = Weight = 100 lbf
we substitute
1500 = 100( h[tex]_f[/tex] - h[tex]_i[/tex] )
h[tex]_f[/tex] - h[tex]_i[/tex] = 1500 / 100
h[tex]_f[/tex] - h[tex]_i[/tex] = 15 ft
given that, elevation of the object; h[tex]_i[/tex] = 30 ft
h[tex]_f[/tex] - 30 ft = 15 ft
h[tex]_f[/tex] = 15 ft + 30 ft
h[tex]_f[/tex] = 45 ft
Therefore, The final elevation is 45 ft
