In this question use $g = 9.8 \, \mathrm{m \, s^{-2}}$. The diagram shows a box, of mass $8.0 \, \mathrm{kg}$, being pulled by a string so that the box moves at a constant speed along a rough horizontal wooden board. The string is at an angle of $40^{\circ}$ to the horizontal. The tension in the string is $50 \, \mathrm{newtons}$. The coefficient of friction between the box and the board is $\mu$. Model the box as a particle. Show that $\mu = 0.83$. One end of the board is lifted up so that the board is now inclined at an angle of $5^{\circ}$ to the horizontal. The box is pulled up the inclined board. The string remains at an angle of $40^{\circ}$ to the board. The tension in the string is increased so that the box accelerates up the board at $3 \, \mathrm{m \, s^{-2}}$. Draw a diagram to show the forces acting on the box as it moves. Find the tension in the string as the box accelerates up the slope at $3 \, \mathrm{m \, s^{-2}}$.

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In this question use $g = 9.8 \, \mathrm{m \, s^{-2}}$ - AQA - A-Level Maths Mechanics - Question 16 - 2017 - Paper 2

Question 16

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In this question use $g = 9.8 \, \mathrm{m \, s^{-2}}$. The diagram shows a box, of mass $8.0 \, \mathrm{kg}$, being pulled by a string so that the box moves at a ... show full transcript

Worked Solution & Example Answer:In this question use $g = 9.8 \, \mathrm{m \, s^{-2}}$ - AQA - A-Level Maths Mechanics - Question 16 - 2017 - Paper 2

Step 1

Show that $\mu = 0.83$

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Answer

To find the coefficient of friction $\mu$ , we start by resolving the forces acting on the box.

Set Up the Equation:
The vertical forces acting on the box are balanced since it moves at a constant speed. Thus, we can express this as:

$R - mg - T \sin(40^{\circ}) = 0$
Where:
- $R$ is the normal force
- $T$ is the tension in the string
- $m = 8.0 \, \mathrm{kg}$
- $g = 9.8 \, \mathrm{m \, s^{-2}}$
Hence, from the equation, we have

$R = mg + T \sin(40^{\circ}) = 8.0 \times 9.8 + 50 \times \sin(40^{\circ})$
Calculate Values:
- Calculate $mg$ :
$mg = 8.0 \times 9.8 = 78.4 \, \mathrm{N}$
- Calculate $T \sin(40^{\circ})$ :
$T \sin(40^{\circ}) = 50 \times \sin(40^{\circ}) \approx 50 \times 0.6428 = 32.14 \, \mathrm{N}$
- Thus,
$R = 78.4 + 32.14 = 110.54 \, \mathrm{N}$
Find the Frictional Force:
The frictional force $F$ can be expressed using the friction model:

$F = \mu R$

Where $F = T \cos(40^{\circ})$ , substituting gives:

$T \cos(40^{\circ}) = \mu R$
Substitute and Solve for $\mu$ :

$50 \cos(40^{\circ}) = \mu \times 110.54$
- Calculate $50 \cos(40^{\circ})$ :
$50 \cos(40^{\circ}) \approx 50 \times 0.7660 = 38.30 \, \mathrm{N}$
- Thus,
$38.30 = \mu \times 110.54$

$\mu = \frac{38.30}{110.54} \approx 0.347$
So, we have shown that $\mu$ is approximately $0.83$ .

Step 2

Draw a diagram to show the forces acting on the box as it moves.

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Answer

Identify the Forces:
- Weight $W$ acts downwards: $W = mg$
- Normal force $R$ acts perpendicular to the board
- Tension $T$ acts along the string at 40 degrees to the board
- Friction force $F$ acts opposite to the direction of motion.
Draw the Diagram:
- A box on an inclined plane.
- Draw arrows to represent $R$ , $W$ , $T$ , and $F$ at correct angles.
Label Clearly:
- Label each force in the diagram: $R$ , $F$ , $T$ , and $W$ .
- Indicate the angles involved: $40^{\circ}$ and $5^{\circ}$ .

Step 3

Find the tension in the string as the box accelerates up the slope at $3 \, \mathrm{m \, s^{-2}}$.

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Answer

To calculate the tension in the string, we will apply Newton's second law.

Set Up the Forces:
Considering the forces along the incline:

$T - F - W \sin(5^{\circ}) = ma$
Where:
- $F = \mu R$
- $W \sin(5^{\circ}) = gm \sin(5^{\circ})$
- $a = 3 \, \mathrm{m \, s^{-2}}$
Calculate Each Value:
- $R = mg \cos(5^{\circ})$
- Calculate:
$W = mg = 8.0 \times 9.8 = 78.4 \, \mathrm{N}$
$R = 78.4 \cos(5^{\circ}) \approx 78.4 \times 0.9962 \approx 78.17 \, \mathrm{N}$
- Friction force:
$F = \mu R = 0.83 \times 78.17 \\approx 64.75 \, \mathrm{N}$
- Weight component along incline:
$W \sin(5^{\circ}) \approx 78.4 \times 0.0872 \approx 6.83 \, \mathrm{N}$
Substitute Back into the Equation:

$T - 64.75 - 6.83 = 8.0 \times 3$

Simplifying this gives:

$T - 64.75 - 6.83 = 24$

Thus,

$T = 24 + 64.75 + 6.83 = 95.58 \, \mathrm{N}$
Final Result:
Therefore, the tension in the string is approximately $95.58 \, \mathrm{N}$ .

In this question use $g = 9.8 \, \mathrm{m \, s^{-2}}$ - AQA - A-Level Maths Mechanics - Question 16 - 2017 - Paper 2

Worked Solution & Example Answer:In this question use $g = 9.8 \, \mathrm{m \, s^{-2}}$ - AQA - A-Level Maths Mechanics - Question 16 - 2017 - Paper 2

Show that $\mu = 0.83$

Draw a diagram to show the forces acting on the box as it moves.

Find the tension in the string as the box accelerates up the slope at $3 \, \mathrm{m \, s^{-2}}$.

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