A crate of mass 18 kg, initially at rest, slides down a frictionless slope from point A to point B - NSC Physical Sciences - Question 5 - 2024 - Paper 1

To find the speed of the crate at point B, we can use the conservation of mechanical energy. At point A, all the energy is potential:

$E_{potential} = mgh = (18 ext{ kg})(9.81 ext{ m/s}^2)(3 ext{ m}) = 529.74 ext{ J}$

At point B, all potential energy is converted to kinetic energy:

E_{kinetic} = rac{1}{2} mv^2

Setting them equal:

529.74 ext{ J} = rac{1}{2} (18 ext{ kg}) v^2

Solving for $v$, we get:

v^2 = rac{529.74 imes 2}{18} $v ext{ at point B} = 7.67 ext{ m/s}$

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.

The work done by the frictional force in moving from point B to point C is equal to the change in kinetic energy.

Using the work-energy theorem:

$W = F imes d = E_{initial} - E_{final}$

Where E_{initial} = rac{1}{2} mv^2 and $E_{final} = 0$, we have:

W = rac{1}{2} (18)(7.67^2) $40.6 imes d = 13.04 ext{ J}$

Solving for $d$ gives the distance travelled by the crate.

The distance moved by the crate after lowering point A will be SMALLER THAN that calculated in QUESTION 5.4 because a lower starting height means lesser potential energy, resulting in less kinetic energy and consequently less distance travelled before coming to rest.