Question

A 1.5-kg mass is projected down a rough circular track with a radius of 2.0 m. The speed of the mass at point A is 3 m/s, and at point B, it is 8.0 m/s. What is the magnitude of the change in mechanical energy done on the system between points A and B by the force of friction? Hint: Find the energy lost due to friction.

Answer 1

The magnitude of the change in mechanical energy done on the system by the force of friction between points A and B is
`\(41.25 \, \text{J}\)`.

The change in mechanical energy is given by the work done by the forces acting on the system. In this case, the force of friction is the primary force doing work, and its work done is equal to the change in mechanical energy.

The work done by the force of friction is given by:

`\[ W_{\text{friction}} = \Delta E_{\text{mechanical}} \]`

The change in mechanical energy can be expressed as the difference in kinetic energy at points A and B:

`\[ \Delta E_{\text{mechanical}} = E_{\text{kinetic, B}} - E_{\text{kinetic, A}} \]`

The kinetic energy of an object is given by the formula:

`\[ E_{\text{kinetic}} = (1)/(2) m v^2 \]`

The following were given:

• Mass, m = 1.5 kg
• Velocity at point A
  `(\(v_{\text{A}}\))` = 3 m/s
• Velocity at point B
  `(\(v_{\text{B}}\))` = 8 m/s

First, let's calculate the kinetic energy at point A:

`\[ E_{\text{kinetic, A}} = (1)/(2) m v_{\text{A}}^2 \]`

`\[ E_{\text{kinetic, A}} = (1)/(2) * 1.5 \, \text{kg} * (3 \, \text{m/s})^2 \]`

`\[ E_{\text{kinetic, A}} = (1)/(2) * 1.5 \, \text{kg} * 9 \, \text{m}^2/\text{s}^2 \]`

= 6.75 J

Next, we calculate the kinetic energy at point B:

`\[ E_{\text{kinetic, B}} = (1)/(2) m v_{\text{B}}^2 \]`

`\[ E_{\text{kinetic, B}} = (1)/(2) * 1.5 \, \text{kg} * (8 \, \text{m/s})^2 \]`

`\[ E_{\text{kinetic, B}} = (1)/(2) * 1.5 \, \text{kg} * 64 \, \text{m}^2/\text{s}^2 \]`

= 48 J

Lastly, we calculate the change in mechanical energy:

`\[ \Delta E_{\text{mechanical}} = E_{\text{kinetic, B}} - E_{\text{kinetic, A}} \]`

`\[ \Delta E_{\text{mechanical}} = 48 \, \text{J} - 6.75 \, \text{J} \]`

= 41.25 J

Therefore, the magnitude of the change in mechanical energy done on the system by the force of friction between points A and B is
`\(41.25 \, \text{J}\)`.