In this question use $g = 9.8 \, \text{m s}^{-2}$. The diagram shows a box, of mass 8.0 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 newtons. The coefficient of friction between the box and the board is $\mu$. Model the box as a particle. (a) Show that $\mu = 0.83$ (b) 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 $3 \, \text{m s}^{-2}$. (b)(i) Draw a diagram to show the forces acting on the box as it moves. (b)(ii) Find the tension in the string as the box accelerates up the slope at $3 \, \text{m s}^{-2}$.

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

Question 16

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

Worked Solution & Example Answer:In this question use $g = 9.8 \, \text{m s}^{-2}$ - 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 resting on the horizontal board.

Identify Forces:
- Weight of the box, $W = mg = 8.0 \, \text{kg} \times 9.8 \, \text{m s}^{-2} = 78.4 \, \text{N}$ .
- Normal reaction force, $R$ .
- Tension in the string, $T = 50 \, \text{N}$ at an angle of $40^{\circ}$ .
Resolve Forces in the Vertical Direction:

Using the equation for vertical forces: $R + T \sin(40^{\circ}) = W$
Substituting known values: $R + 50 \sin(40^{\circ}) = 78.4 \Rightarrow R = 78.4 - 50 \sin(40^{\circ})$
Calculating: $R \approx 78.4 - 50 \times 0.6428 = 78.4 - 32.14 = 46.26 \, \text{N}$
Resolve Forces in the Horizontal Direction:

The horizontal forces must balance since the box is moving at constant speed: $T \cos(40^{\circ}) = F_{friction} = \mu R$
Substituting for tension: $50 \cos(40^{\circ}) = \mu \times 46.26$
Calculate and Solve for $\mu$ :

$\mu = \frac{50 \cos(40^{\circ})}{46.26} \approx \frac{38.85}{46.26} \approx 0.839 \approx 0.83.$
Therefore, we have shown that $\mu = 0.83$ .

Step 2

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

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Answer

To draw the forces acting on the box:

Represent the box on an inclined plane.
Draw the weight vector $W$ acting downward (vertically down).
Draw the normal force $R$ , perpendicular to the surface of the inclined board.
Draw the tension force $T$ at an angle of $40^{\circ}$ to the board.
Draw the frictional force $F_{friction}$ acting down the slope, opposite to the direction of motion.
Label all forces clearly.

Step 3

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

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Answer

To find the tension $T$ in the string when the box accelerates up the incline:

Draw a Free Body Diagram (FBD): Include all forces acting on the box as described above.
- Weight: $W = 78.4 \, \text{N}$
- Normal Reaction: $R$
- Tension: $T$
- Friction: $F_{friction} = \mu R$
Resolve Forces:

In the direction parallel to the incline: $T - F_{friction} - W \sin(5^{\circ}) = ma$ Substitute known values: $T - \mu R - 78.4 \sin(5^{\circ}) = 8.0 imes 3$ where $R$ needs to be calculated at this incline: $R = mg \cos(5^{\circ}) = 8.0 \times 9.8 \times \cos(5^{\circ}) \approx 77.92 \, \text{N}$ Thus, $F_{friction} = \mu R = 0.83 \times 77.92 \approx 64.67 \, \text{N}$
Plug into the equation:

$T - 64.67 - 78.4 \times 0.0872 = 24\rightarrow \text{substituting for acceleration part}$

Calculate:
- $78.4 \times 0.0872 \approx 6.84$
- Thus, plugging back:

ightarrow T \approx 95.51 , \text{N}$$ The final computed tension in the string is approximately $T \approx 95.5 \, \text{N}$ .

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

Worked Solution & Example Answer:In this question use $g = 9.8 \, \text{m s}^{-2}$ - 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 $3 \, \text{m s}^{-2}$.

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