Final answer:
The work done by the applied force (tension in the rope) on the 2.0 kg bucket is 237.6 J, and the work done by gravity is -194.04 J, calculated over a displacement of 9.9 m resulting from the applied acceleration over 3.0 s.
Step-by-step explanation:
To calculate the work done by each force acting on a 2.0 kg bucket that is accelerated from rest at 2.2 m/s2 for 3.0 s, we first need to determine the displacement and the forces involved. The bucket is subjected to two forces: the force of gravity (weight) and the tension in the rope (applied force). The work done by gravity will be negative since it acts in the opposite direction of the displacement.
The displacement (s) during acceleration can be calculated using the kinematic equation:
s = ut + ½at2
where:
- u = initial velocity = 0 m/s (starting from rest)
- a = acceleration = 2.2 m/s2
- t = time = 3.0 s
Plugging in the values we get:
s = 0 + ½(2.2 m/s2)(3.0 s)2 = 9.9 m
The work done by the applied force is:
W = Fapplied × s × cos(θ)
where θ is the angle between the force and the displacement which is 0° since the rope pulls straight up. The force applied by the rope can be found using Newton's second law (F = ma), considering the force needed to overcome gravity and to provide the acceleration.
Fapplied = m(g + a) = (2.0 kg)(9.8 m/s2 + 2.2 m/s2) = 24 N
Therefore, the work done by the tension in the rope is:
Wapplied = (24 N)(9.9 m) = 237.6 J
The work done by the gravitational force (Wgravity) is:
Wgravity = mgh = (2.0 kg)(9.8 m/s2)(-9.9 m) = -194.04 J (negative because gravity is opposite to the displacement)