Assume that the amplifiers on the stage operate at a combined power of 5,000 Watts, and that 80% of this is converted to sound.
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(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to
where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and is the variation of height. Substituting the numbers into the formula, we find
(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:
However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.
(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):
1) Time in the air: 0.78 s
The motion of the ball is a projectile motion, which consists of two independent motions:
- A horizontal motion with constant horizontal velocity
- A vertical motion with constant downward acceleration of
(acceleration of gravity)
The initial vertical velocity is
where the negative sign means the direction is downward.
The vertical position of the ball is given by
where
h = 4.50 m is the initial heigth of the ball when it starts falling down
The ball reaches the ground when y = 0, so we have:
This is a second-order equation; solving for t, we get
t = -1.18 s
t = 0.78 s
We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.
2) Horizontal distance: 1.83 m
For this second part of the problem, we just have to consider the horizontal motion of the ball.
As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by
where u = 3.05 m/s is the initial speed and the angle of projection.
For a uniform motion, we can use the following relationship between distance covered and velocity:
and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground: