A particle, under the action of two constant forces, is moving across a perfectly smooth horizontal surface at a constant speed of 10 m s⁻¹. The first force acting on the particle is $(400 ext{ N})$. The second force acting on the particle is $( ext{ }(180) ext{ N})$. Find the value of $p$. Circle your answer.

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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 m s⁻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 m s⁻¹. The first force acting ... 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 m s⁻¹ - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 2

Step 1

Find the net force acting on the particle

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Answer

Since the particle is moving at a constant speed, the net force acting on it is zero. Therefore, the forces must balance each other. We denote the two forces as follows:

First force: $F_1 = 400 ext{ N}$ (acting in one direction)
Second force: $F_2 = p ext{ N}$ (acting in the opposite direction)

Setting them equal:
$F_1 - F_2 = 0$
This gives us:
$400 - p = 0$

Step 2

Solve for p

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Answer

Rearranging the equation results in: $p = 400 ext{ N}$

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