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 resistive force of 0.3R newtons, where R is the normal force, as shown in the diagram. Take the acceleration g due to gravity to be 10 m/s². (i) By resolving the forces vertically, show that R = 400. (ii) Show that the net force horizontally is approximately 53.2 newtons. (iii) Find the velocity of the box after the first three seconds.

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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.

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Answer

To solve for the normal force R, we start by applying the equation of motion in the vertical direction:

$ext{Sum of vertical forces} = 0$

The upward forces include the vertical component of the pulling force, which is given by:

$200 imes ext{sin}(30°)$

The downward forces include the weight of the box, which can be calculated as:

$ext{Weight} = 50 imes 10 = 500 ext{ N}$

Setting up the equation:

$200 imes ext{sin}(30°) + R = 500$

Calculating the component:

$200 imes 0.5 = 100$

Thus:

$100 + R = 500$

Rearranging gives:

$R = 500 - 100 = 400 ext{ N}$

Step 2

Show that the net force horizontally is approximately 53.2 newtons.

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Answer

To find the net horizontal force, we first calculate the horizontal component of the 200 N force:

$F_{x} = 200 imes ext{cos}(30°)$

Calculating:

$F_{x} = 200 imes rac{ ext{sqrt}(3)}{2} = 100 ext{sqrt}(3) ext{ N} \\ \approx 173.2 ext{ N}$

Next, we consider the resistive force, which is given as 0.3R:

$ext{Resistive force} = 0.3R = 0.3 imes 400 = 120 ext{ N}$

Then, the net horizontal force is:

$F_{net} = F_{x} - ext{Resistive force}$

Substituting in the values:

$F_{net} = 173.2 - 120 = 53.2 ext{ N}$

Step 3

Find the velocity of the box after the first three seconds.

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Answer

To find the velocity after three seconds, we first calculate the acceleration of the box using Newton's second law:

$F = ma$

From part (ii), we found that the net force is 53.2 N, and the mass of the box is 50 kg. Thus:

$a = \frac{F_{net}}{m} = \frac{53.2}{50} = 1.064 ext{ m/s}^2$

Now we use the formula for velocity when starting from rest:

$v = u + at$

Since the initial velocity, $u$ , is 0 (the box starts from rest), we find:

$v = 0 + (1.064)(3) = 3.192 ext{ m/s}$

Therefore, the velocity of the box after three seconds is approximately 3.192 m/s.

A 50-kilogram box is initially at rest - HSC - SSCE Mathematics Extension 2 - Question 12 - 2020 - Paper 1

Worked Solution & Example Answer:A 50-kilogram box is initially at rest - HSC - SSCE Mathematics Extension 2 - Question 12 - 2020 - Paper 1

By resolving the forces vertically, show that R = 400.

Show that the net force horizontally is approximately 53.2 newtons.

Find the velocity of the box after the first three seconds.

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