Answer: 219,344.5 j
Explanation:
from the question we have the following parameters
mass of barge (m) = 5 x 10^4 kg
angle of ropes (θ) = 45 degrees
force by each mule (F1) = 1.10 kN = 1100 N
force by each mule (F2) = 1.10 kN = 1100 N
distance (s) = 141 m
work done by the mules = force x cos θ x distance
since we are calculating the force of both mules we have to both mules before we can use in the equation
force = 1100 + 1100 = 2200 N
work done = 2200 x cos (45) x 141
work done = 219,344.5 j
Let u = upward initial speed of the flea.
At the maximum height of h = 0.550 m, the vertical velocity is v = 0.
Therefore, from the formula
v² = u² - 2gh,
obtain
0 = u² - 2*(9.8 m/s²)*(0.55 m)
u² = 10.78 m²/s²
u = 3.283 m/s
Answer: The initial upward speed is 3.28 m/s (nearest thousandth)
Answer:
(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?
No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum
(b)Is it possible for only one to be at rest after the collision?
Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.
Writing the acceleration as a function of time:
a(t) = 1 + 3√t
Integrating acceleration, we obtain velocity:
v(t) = t + 2(t)^(3/2) + c;
object at rest so velocity at t = 0 is 0 so c = 0.
v(t) = t + 2(t)^(3/2)
Integrating velocity to obtain an equation for displacement:
d(t) = t²/2 + 4/5 t^(5/2) + c
Applying limits from t = 0 to t = 9
d = 9²/2 + 4/5 9^(5/2)
d = 234.9 m
