A person pushes a lawn mower of mass 15 kg at a constant speed in a straight line over a flat grass surface with a force of 90 N - NSC Physical Sciences - Question 2 - 2019 - Paper 1

The lawn mower is in equilibrium because it is moving at a constant speed in a straight line. The net force acting on it is zero; hence the forces balance out. According to Newton's first law, an object will remain at rest or in uniform motion unless acted upon by a net external force.

Using the equilibrium condition:

The horizontal forces must balance out:

$F - f = 0$

Here, F = 90 ext{ N}\ (the applied force), and $f$ is the frictional force. Therefore:

$f = 90 ext{ N} - F_{x}$

Where the component of the applied force in the x-direction is:

$F_{x} = F imes ext{cos}(40°)$

determining:

$F_{x} = 90 imes ext{cos}(40°) \\ ext{= } 68.94 ext{ N}$

Thus,

$f = 90 ext{ N} - 68.94 ext{ N} = 21.06 ext{ N}$

First, calculate the acceleration ($a$):

$a = \frac{(v_f - v_i)}{t} = \frac{(2 ext{ m/s} - 0 ext{ m/s})}{3 ext{ s}} = \frac{2}{3} ext{ m/s}^2 \\ a = 0.67 ext{ m/s}^2$

Now apply Newton's second law:

$F_{net} = m \cdot a$

Where mass $m = 15 ext{ kg}$:

$F_{net} = 15 ext{ kg} \cdot 0.67 ext{ m/s}^2 = 10.05 ext{ N}$

Now account for friction:

Thus, the total force ($F$) required is:

$F = F_{net} + f = 10.05 ext{ N} + 21.06 ext{ N} = 31.11 ext{ N}$

Using the formula for gravitational force:

$F = \frac{G \cdot m_1 \cdot m_2}{r^2}$

Where:

• $F = 20 ext{ N}$ (the gravitational force)
• $m_1 = 10 ext{ kg}$ (the mass of the object)
• $r = 6 \times 10^5 ext{ m}$ (the radius of the planet)
• $G = 6.67 \times 10^{-11} \text{ N m}^2/ ext{kg}^2$ (gravitational constant)

Rearranging the formula to solve for the mass of the planet ($m_{planet}$):

$m_{planet} = \frac{F \cdot r^2}{G \cdot m_1}$

Substituting the values:

$m_{planet} = \frac{20 ext{ N} \cdot (6 \times 10^5 ext{ m})^2}{6.67 \times 10^{-11} \cdot 10 ext{ kg}}$

Calculating:

$m_{planet} = 1.08 \times 10^{22} ext{ kg}$