An object of mass 1.8 kg slides down a rough curved track and passes point A, which is 1.5 m above the ground, at a speed of 0.95 m·s⁻¹ - NSC Physical Sciences - Question 5 - 2019 - Paper 1

The gravitational force is the conservative force acting on the object.

NO. Mechanical energy is not conserved due to the presence of friction along the track, which transforms some mechanical energy into thermal energy.

The gravitational potential energy (PE) at point A can be calculated using the formula:

$PE = mgh$

where:
$m = 1.8 \, kg$
$g = 9.81 \, m/s^2$
$h = 1.5 \, m$

Thus,

$PE = 1.8 imes 9.81 imes 1.5 = 26.46 \, J$

Using the work-energy principle, the work done by friction (W_f) can be calculated as follows:

Initial mechanical energy at point A (E_A) is the sum of potential and kinetic energy:
$E_A = PE + KE = 26.46 \, J + \frac{1}{2} m v^2 = 26.46 + \frac{1}{2} (1.8)(0.95^2)$

Calculate the kinetic energy (KE):
$KE = \frac{1}{2} (1.8)(0.95^2) = 0.82 \, J$

Thus,
$E_A = 26.46 + 0.82 = 27.28 \, J$

At point B, since the object is at the bottom, its potential energy is zero and its kinetic energy is:
$KE_B = \frac{1}{2} mv^2 = \frac{1}{2} (1.8)(4^2) = 14.4 \, J$

Now, equate the energies considering work done against friction:
$W_f = E_A - E_B = 27.28 - 14.4 = -12.88 \, J$

The net work done (W_net) on the object as it slides from point B to point C is zero, as surface BC is frictionless. Therefore, the net work done is given by:
$W_{net} = 0 \, J$