A 2 kg box is released from rest at point P, 5 m above the ground - NSC Physical Sciences - Question 5 - 2021 - Paper 1

Using the principle of conservation of mechanical energy:

At point P:

• Potential Energy (PE) = mgh = (2 ext{ kg})(9.81 ext{ m/s}²)(5 ext{ m}) = 98.1 ext{ J}
• Kinetic Energy (KE) = 0 ext{ J} (since it starts from rest)

At point Q:

• Total mechanical energy = PE + KE = 98.1 ext{ J} + 0 = 98.1 ext{ J}

• At point Q, the potential energy (h) becomes 0, thus:

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

  98.1 = rac{1}{2} (2)v^2

  $v^2 = 98.1 ext{ J}$

  v \, \approx 9.9 \text{ m/s} $$

At point R, the box moves with a speed of 4 m·s⁻¹. We can use energy principles to analyze the energy lost due to friction as it moves from Q to R.

Using the work-energy principle:

The work done against friction (W_f) can be determined by the change in kinetic energy:

Initial KE at point Q:

• KE_i = rac{1}{2} mv^2 = rac{1}{2} (2) (9.9)^2 = 19.8 ext{ J}

Final KE at point R:

• KE_f = rac{1}{2} mv^2 = rac{1}{2} (2) (4)^2 = 16 ext{ J}

The change in kinetic energy:

• $riangle KE = KE_f - KE_i = 16 ext{ J} - 19.8 ext{ J} = -3.8 ext{ J}$

The average frictional force (F_f) can be calculated using:

• F_f = rac{- riangle KE}{d} = rac{-(-3.8)}{10 ext{ m}} = 0.38 ext{ N}

To calculate the change in kinetic energy after striking the barrier:

Let initial kinetic energy just before impact at R be:

• KE_i = rac{1}{2} mv^2 = rac{1}{2} (2) (4)^2 = 16 ext{ J}

After the collision, the box experiences an impulse of -14 N·s, leading to a change in velocity:

Using:

• $F riangle t = m(v_f - v_i)$ -14 = 2(v_f - 4)
  Thus, the final velocity after the collision:
• $v_f = 4 - 7 = -3$

Now, the final kinetic energy (KE_f) after striking the barrier:

• KE_f = rac{1}{2} (2)(-3)^2 = rac{1}{2} (2)(9) = 9 ext{ J}

Finally, change in kinetic energy:

• $riangle KE = KE_f - KE_i = 9 ext{ J} - 16 ext{ J} = -7 ext{ J}$