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A 50-kilogram box is initially at rest - HSC - SSCE Mathematics Extension 2 - Question 12 - 2020 - Paper 1
Question 12
A 50-kilogram box is initially at rest. The box is pulled along the ground with a force of 200 newtons at an angle of 30° to the horizontal. The box experiences a re... show full transcript
Worked Solution & Example Answer:A 50-kilogram box is initially at rest - HSC - SSCE Mathematics Extension 2 - Question 12 - 2020 - Paper 1
Step 1
By resolving the forces vertically, show that R = 400.
Answer
To resolve the forces vertically, we can apply the following equilibrium equation:
The vertical forces acting on the box include the weight of the box and the vertical component of the pulling force:
egin{align*} 200 imes ext{sin}(30°) + R = 50 imes 10 \ \Rightarrow R = 50 imes 10 - 200 imes \frac{1}{2} \ \Rightarrow R = 500 - 100 \ \Rightarrow R = 400 ext{ newtons.} \end{align*}
Step 2
Show that the net force horizontally is approximately 53.2 newtons.
Answer
To find the net horizontal force, we analyze the horizontal components of the forces acting on the box:
The horizontal component of the pulling force and the resistive force are as follows:
egin{align*} F_{horizontal} = 200 imes ext{cos}(30°) - 0.3R \ \Rightarrow F_{horizontal} = 200 \times \frac{\sqrt{3}}{2} - 0.3 \times 400 \ \Rightarrow F_{horizontal} = 100\sqrt{3} - 120 \ \Rightarrow F_{horizontal} \approx 173.2 - 120 \ \Rightarrow F_{horizontal} \approx 53.2 ext{ newtons.} \end{align*}
Step 3
Find the velocity of the box after the first three seconds.
Answer
Using the second law of motion, we can determine the acceleration of the box:
egin{align*} F_{net} = ma \ 53.2 = 50a \ \Rightarrow a = \frac{53.2}{50} \approx 1.064 ext{ m/s².} \end{align*}
To find the velocity after 3 seconds, we use the formula:
egin{align*} v = u + at \ = 0 + 1.064 \times 3 \ = 3.192 ext{ m/s.} \end{align*}