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

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

$mgh_A = \frac{1}{2}mv^2 + mgh_B$

Where:

• $m = 4 \, kg$ (mass of the box)
• $h_A = 7 \, m$ (height at A)
• $h_B = 7 - 3 \sin(60^{\circ}) \approx 7 - 2.598 \approx 4.402 \, m$ (height at B)
• $g = 9.8 \, m/s^2$ (acceleration due to gravity)

Plugging in these values, we find:

$4 imes 9.8 imes (7) = \frac{1}{2} \times 4 \times v^2 + 4 \times 9.8 \times (4.402)$

Solving for $v$ gives: $v \approx 7.14 \, m/s$

The work done by a net force is equal to the change in kinetic energy of the object.

For the free-body diagram, include:

• Weight ($W$) acting downwards,
• Normal force ($N$) acting perpendicular to the surface,
• Frictional force ($f$) acting opposite to the direction of motion.

Using the Work-Energy principle:

$W_{net} = \Delta KE$

Where:

• $W_{net} = f \cdot d$ (net work done, we need to calculate the distance from B to C)
• $\Delta KE = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_C^2$
• Let’s use $d = 4 \, m$

Assuming the speed at position C ($v_C$) is less than 3 m/s (due to friction), use: $f = W_{net} / d$

The calculated result reveals that frictional force $f \approx 54.94 \, N$.

Remains the same.