A 4 kg box at point A above the horizontal is released and slides down 7 m from A to C on an incline - NSC Physical Sciences - Question 4 - 2017 - Paper 1

To calculate the speed of the box at position B, we can use the principle of conservation of energy:

$E_{mec, A} = E_{mec, B}$

Where:

• At point A:
  • Potential Energy (PE) = mgh = 4 imes 9.8 imes (7 imes ext{sin} ext{ } 60°)
• At point B:
  • Total Energy = PE + KE = PE + \frac{1}{2}mv^2

By equating these two expressions, we can solve for the speed, $v$ at point B.

Work done by a net force is equal to the change in kinetic energy, which states that the work required to change the energy in any system must equal that change.

The free-body diagram would show:

• Weight (W) acting downwards.
• Normal force (N) acting perpendicular to the surface.
• Frictional force (f) acting opposite to the direction of motion. The detailed representation depends on the specific scenario with labeled forces.

Using the Work-Energy Theorem:

$W_{net} = \Delta KE$

Calculating the work done against friction:

• Total work = (Frictional force imes distance)
• $f = \frac{W_{net}}{d}$

Finally, substituting the values to find the frictional force.

The coefficient of kinetic friction remains the same as the frictional force depends on the nature of the surfaces in contact, not the angle of the incline.