A particle, under the action of two constant forces, is moving across a perfectly smooth horizontal surface at a constant speed of 10 ms$^{-1}$. The first force acting on the particle is $(400 + 180)\,\text{N}$. The second force acting on the particle is $(\pi - 180)\,\text{N}$. Find the value of $p$. Circle your answer. -400 -390 -400 400

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A particle, under the action of two constant forces, is moving across a perfectly smooth horizontal surface at a constant speed of 10 ms$^{-1}$ - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 2

Question 12

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A particle, under the action of two constant forces, is moving across a perfectly smooth horizontal surface at a constant speed of 10 ms$^{-1}$. The first force act... show full transcript

Worked Solution & Example Answer:A particle, under the action of two constant forces, is moving across a perfectly smooth horizontal surface at a constant speed of 10 ms$^{-1}$ - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 2

Step 1

Calculate the resultant force

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Answer

To find the constant speed, the net force acting on the particle must be zero. Thus:

$F_{net} = F_{1} - F_{2} = 0$

Where:

$F_{1} = 400 + 180 = 580 \text{ N}$
$F_{2} = \pi - 180 \text{ N}$

Setting up the equation:

$580 - (\pi - 180) = 0$

Step 2

Solve for p

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Answer

Rearranging the equation gives:

$580 = \pi - 180$

Solving for $\pi$ :

$\pi = 580 + 180 = 760$

The correct value of $p$ is thus:

$p = 400$

Therefore, circle the answer -400.

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