2.1 Draw a labelled free-body diagram for the lawn mower - NSC Physical Sciences - Question 2 - 2019 - Paper 1

The lawn mower is in equilibrium because it moves at a constant speed in a straight line. According to Newton's First Law, this indicates that the net force acting on the lawn mower is zero. The applied force and the frictional force are equal in magnitude but opposite in direction.

Using the equations from the free-body diagram, we need to set the net force in the horizontal direction to zero:

1. Identify the forces:

   • Horizontal component of the applied force:

     $F_{A_x} = F_A imes ext{cos}(40°) = 90 ext{ N} imes ext{cos}(40°)$

2. Calculate:

   • $F_{A_x} = 90 ext{ N} imes 0.766 = 68.94 ext{ N}$

3. Set the frictional force equal to the horizontal applied force:

   $f = F_{A_x} = 68.94 ext{ N}$

To find the required force,

1. Calculate the mass using the lawn mower's weight:

   m = rac{F_g}{g} = rac{90 ext{ N}}{9.81 ext{ m·s}^{-2}} \approx 9.17 ext{ kg}

2. Use Newton's Second Law:

   $F_{net} = m imes a$

3. Calculate acceleration:

   a = rac{ ext{final velocity} - ext{initial velocity}}{ ext{time}} = rac{2 ext{ m·s}^{-1} - 0 ext{ m·s}^{-1}}{3 ext{ s}} = rac{2}{3} ext{ m·s}^{-2}

4. Substitute to find the net force:

   F_{net} = 9.17 ext{ kg} imes rac{2}{3} ext{ m·s}^{-2} \approx 6.11 ext{ N}

5. Calculate the total required force by adding the frictional force:

   • Total force = Frictional force + Required net force

   $F = 68.94 ext{ N} + 6.11 ext{ N} \approx 75.05 ext{ N}$

Using the formula for gravitational force:

1. The weight of the mass on planet Y:

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

Substituting the known values:

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

Solving for the mass of planet Y:

$m_{planet} = \frac{20 ext{ N} \cdot (6 \times 10^5 m)^2}{(6.67 \times 10^{-11} imes 10 ext{ kg})} \approx 1.08 \times 10^{22} ext{ kg}$