A demolition ball is used by a crane to break the wall of a building - NSC Physical Sciences - Question 5 - 2021 - Paper 1

To find the average power $ext{P}$, we use the formula:

ext{P} = rac{W}{ riangle t}

Where:

• $ext{W}$ is the work done in lifting the ball,
• $riangle t$ is the time taken.

Calculating work done, we have:

$W = mgh$ Where:

• $m = 1250 ext{ kg}$,
• $g = 9.8 ext{ m/s}^2$,
• $h = 5.8 ext{ m}$.

Thus,

$W = 1250 imes 9.8 imes 5.8 = 72 250 ext{ J}$

The time is given as:

• $riangle t = 60 ext{ s}$.

So, substituting back:

ext{P} = rac{72 250}{60} = 1 204.17 ext{ W}

The average power dissipated by the crane is approximately 1 204 W.

A conservative force is defined as a force for which the work done in moving an object between two points does not depend on the path taken. Examples include gravitational and electrostatic forces.

The force exerted by the wall on the ball is a NON-CONSERVATIVE force. This is due to the fact that the work done by the wall's force depends on the displacement of the ball into the wall, and some energy is dissipated as sound or deformation.

During the downward swing of the demolition ball, gravitational potential energy is converted into kinetic energy. As the ball descends, its height decreases, resulting in a decrease in potential energy and an increase in its kinetic energy.

To calculate the average force exerted during the collision, we can use the work-energy principle:

$W = F imes d$ Where:

• $W$ is the work done on the ball,
• $F$ is the average force,
• $d$ is the distance moved into the wall (0.25 m).

From the previous calculations, the energy lost is equal to the kinetic energy just before hitting the wall:

K_i = rac{1}{2} mv^2

Before impact, where $v$ can be calculated from potential energy: Using $h = 5.8 ext{ m}$ as initial height:

$K_i = mgh = 1250 imes 9.8 imes 5.8$

Now, the potential energy becomes:

$K_i = 72 250 ext{ J}.$ The work done in moving into the wall (force applied) equals work done:

$F imes 0.25 = 72 250$

Solving for $F$:

F = rac{72 250}{0.25} = 289 000 ext{ N}.

Thus, the magnitude of the average force exerted on the ball is approximately 289 000 N.