A particle P of mass 0.5 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = 3/4. The particle is held at rest on the plane by the action of a force of magnitude 4 N acting up the plane in a direction parallel to a line of greatest slope of the plane, as shown in Figure 2. The particle is on the point of slipping up the plane. (a) Find the coefficient of friction between P and the plane. The force of magnitude 4 N is removed. (b) Find the acceleration of P down the plane.

A-Level Edexcel Maths Mechanics Forces: A particle P of mass 0.5 kg is o

A particle P of mass 0.5 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = 3/4 - Edexcel - A-Level Maths Mechanics - Question 4 - 2006 - Paper 1

Question 4

A particle P of mass 0.5 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = 3/4. The particle is held at rest on the plane by the action ... show full transcript

Worked Solution & Example Answer:A particle P of mass 0.5 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = 3/4 - Edexcel - A-Level Maths Mechanics - Question 4 - 2006 - Paper 1

Step 1

Find the coefficient of friction between P and the plane.

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Answer

To find the coefficient of friction (μ) between particle P and the plane, we start by analyzing the forces acting on the particle.

Forces acting on the particle:
- Weight (W) acting downwards:
  $W = mg = 0.5 imes 9.8 = 4.9 ext{ N}$
- The component of weight acting down the slope can be calculated as:
  $W_{ ext{down}} = W imes ext{sin} α = 4.9 imes rac{3}{5} = 2.94 ext{ N}$
- The component of weight perpendicular to the slope is:
  $W_{ ext{perpendicular}} = W imes ext{cos} α = 4.9 imes rac{4}{5} = 3.92 ext{ N}$
- Normal reaction force (R) on the plane:
  $R = W_{ ext{perpendicular}} = 3.92 ext{ N}$
Apply equilibrium conditions: The particle is on the point of slipping, thus the forces parallel to the slope are in equilibrium: $F_{ ext{applied}} = W_{ ext{down}} + F_{ ext{friction}}$ Where,
$F_{ ext{friction}} = μR$
- Plugging in the forces gives: $4 = 2.94 + μ(3.92)$
- Rearranging to solve for μ yields: $μ(3.92) = 4 - 2.94$ $μ(3.92) = 1.06$ $μ = rac{1.06}{3.92} = 0.27$

Thus, the coefficient of friction between P and the plane is 0.27.

Step 2

Find the acceleration of P down the plane.

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Answer

After the force of magnitude 4 N is removed, we analyze the forces to find the acceleration of P down the slope:

Summarize the forces acting on P: With the applied force gone, the only forces acting along the slope are the gravitational force pulling it down the plane and the frictional force opposing that motion.
- The gravitational component down the slope was previously calculated as:
  $W_{ ext{down}} = 2.94 ext{ N}$
- The frictional force will now be:
  $F_{ ext{friction}} = μR = 0.27 imes 3.92 = 1.06 ext{ N}$
Apply Newton's second law (F = ma): The net force acting down the slope can be calculated as:
$F_{ ext{net}} = W_{ ext{down}} - F_{ ext{friction}}$ $F_{ ext{net}} = 2.94 - 1.06 = 1.88 ext{ N}$
Calculate the mass of P: The mass (m) of P is 0.5 kg.
Find the acceleration (a): Using Newton's second law:
$F_{ ext{net}} = ma$ Rearranging gives: $a = rac{F_{ ext{net}}}{m} = rac{1.88}{0.5} = 3.76 ext{ m/s}^2$

Thus, the acceleration of P down the plane is approximately 3.76 m/s².

A particle P of mass 0.5 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = 3/4 - Edexcel - A-Level Maths Mechanics - Question 4 - 2006 - Paper 1

Worked Solution & Example Answer:A particle P of mass 0.5 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = 3/4 - Edexcel - A-Level Maths Mechanics - Question 4 - 2006 - Paper 1

Find the coefficient of friction between P and the plane.

Find the acceleration of P down the plane.

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