(ii) linear motion.
(b) Describe a laboratory experiment to determine the density of an irregularly shaped solid.
(d) Explain the term inertia.
The diagram above illustrates a body of mass 5.0 kg being pulled by a horizontal force F. If the body accelerates at 2.0 ms\(^{-2}\) and experiences a frictional force of 5 N, calculate the:
(iii) coefficient of kinetic friction. [ g = 10 ms\(^{-2}\)]
(a) Examples of motion
(i) Rotational motion: a spinning wheel or tyre; the blades of a rotating fan; a spinning top.
(ii) Linear motion: a car moving along a straight road; a ball falling freely under gravity; an object sliding along a straight track.
(b) Experiment to determine the density of an irregularly shaped solid
- Find the mass \(m\) of the solid using a beam or electronic balance.
- Partly fill a measuring cylinder with water and note the initial volume reading \(V_1\).
- Tie the solid with a thin thread and lower it gently until it is fully immersed in the water. Note the new volume reading \(V_2\).
- The volume of the solid is \(V = V_2 - V_1\) (volume of water displaced).
- Calculate the density from \(\rho = \dfrac{m}{V_2 - V_1}\).
(c) Newton's second law of motion
The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(d) Inertia
Inertia is the property of a body by virtue of which it resists any change to its state of rest or of uniform motion in a straight line. The greater the mass of a body, the greater its inertia.
(e) Calculations (mass \(m = 5.0\) kg, acceleration \(a = 2.0\ \text{m s}^{-2}\), friction \(f = 5\) N, \(g = 10\ \text{m s}^{-2}\); F is horizontal)
(i) Net force
\[ F_{net} = ma = 5.0 \times 2.0 = 10\ \text{N} \]
(ii) Magnitude of F
The horizontal pull provides the net force after overcoming friction:
\[ F - f = F_{net} \;\Rightarrow\; F = F_{net} + f = 10 + 5 = 15\ \text{N} \]
(iii) Coefficient of kinetic friction
The normal reaction equals the weight: \(R = mg = 5.0 \times 10 = 50\ \text{N}\).
\[ \mu = \frac{f}{R} = \frac{5}{50} = 0.1 \]