(a)
Force per unit mass (on an object in a gravitational field).
Gravitational field strength, denoted by $g$, defines the intensity of a gravitational field. It is formally defined as the gravitational force ($W$) exerted on an object per unit of its mass ($m$): $g = \frac{W}{m}$.
(b)(i)
$16\text{ N}$
First, find the mass on Earth: $m = \frac{W}{g} = \frac{42}{9.8} \approx 4.286\text{ kg}$. Weight on Mars is $W = m \times g_{\text{Mars}} = 4.286 \times 3.7 = 15.86\text{ N} \approx 16\text{ N}$.
(b)(ii)
$1.6 \times 10^{20}\text{ m}^3$
Using $\rho = \frac{m}{V}$, then $V = \frac{m}{\rho}$. $V = \frac{6.4 \times 10^{23}}{3900} \approx 1.641 \times 10^{20}\text{ m}^3$.
(c)(i)
An arrow drawn parallel to the driving force, pointing to the right, and labelled $30\text{ N}$. Since the buggy moves at constant speed, the resultant force is zero; thus resistive forces must equal the driving force in the opposite direction.
(c)(ii)
1. The resultant force increases (it is no longer zero).
2. Resultant force = driving force $-$ resistive force.
