Question

An object with mass 60 kg moved in outer space. When it was at location <13, -25, -4> its speed was 4.5 m/s. A single constant force <240, 480, -170> N acted on the object while the object moved from location <13, -25, -4> m to location <16, -17, -7> m. Then a different single constant force <80, 230, 180> N acted on the object while the object moved from location <16, -17, -7> m to location <22, -23, -1> m. What is the speed of the object at this final location?

Answer 1

let
m = 60 kg
vi = 4.5 m/s
F1 = (240 i + 480 j - 170 k) N
delta_x1 = (16 - 13)i + (-17 - (-25))j + (-7 - (-4))j

= (3i + 8j - 3k) m

F2 = (80i + 230j + 180 k) N
delta_x2 = (22 - 16)i + (-23 - (-17))j + (-1 - (-7))k

= (6i - 6j + 6k) m

total workdone, Wnet = W1 + W2

= F1.delta_x1 + F2.delta_x2

= (240 i + 480 j - 170 k).(3i + 8j - 3k) + (80i + 230j + 180 k).(6i - 6j + 6k)

= 240*3 + 480*8 + (-170)*(-3) + 80*6 + 230*(-6) + 180*6

= 5250 J

now use, Work-energy theorem,

Wnet = change in kinetic energy

Wnet = (1/2)*m*(vf^2 - vi^2)

2*Wnet/m = vf^2 - vi^2

vf^2 = vi^2 + 2*Wnet/m

vf = sqrt(vi^2 + 2*Wnet/m)

= sqrt(4.5^2 + 2*5250/60)

= 14 m/s <<<<<<<<<<------------------------Answer