A particle of mass 0.8 kg is held at rest on a rough plane. The plane is inclined at 30° to the horizontal. The particle is released from rest and slides down a line of greatest slope of the plane. The particle moves 2.7 m during the first 3 seconds of its motion. (a) the acceleration of the particle, (b) the coefficient of friction between the particle and the plane. The particle is now held on the same rough plane by a horizontal force of magnitude X newtons, acting in a plane containing a line of greatest slope of the plane, as shown in Figure 3. The particle is in equilibrium and on the point of moving up the plane. (c) Find the value of X.

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A particle of mass 0.8 kg is held at rest on a rough plane - Edexcel - A-Level Maths Mechanics - Question 5 - 2010 - Paper 1

Question 5

A particle of mass 0.8 kg is held at rest on a rough plane. The plane is inclined at 30° to the horizontal. The particle is released from rest and slides down a line... show full transcript

Worked Solution & Example Answer:A particle of mass 0.8 kg is held at rest on a rough plane - Edexcel - A-Level Maths Mechanics - Question 5 - 2010 - Paper 1

Step 1

a) the acceleration of the particle

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Answer

To find the acceleration of the particle, we can use the formula:

$s = ut + \frac{1}{2}at^2$

Where:

$s = 2.7$ m
$u = 0$ (initial velocity, since it starts from rest)
$t = 3$ s
$a$ is the acceleration.

Substituting the values, we rearrange the equation to solve for $a$ :

$2.7 = 0 + \frac{1}{2}a(3^2) \implies 2.7 = \frac{1}{2} \times a \times 9 \implies a = \frac{2.7 \times 2}{9} = 0.6 \, \text{m/s}^2$

Thus, the acceleration of the particle is $0.6 \, \text{m/s}^2$ .

Step 2

b) the coefficient of friction between the particle and the plane

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The forces acting on the particle can be analyzed. The weight components along and perpendicular to the plane are:

Weight perpendicular: $R = mg \cos(30°)$
Weight parallel: $mg \sin(30°)$

Where:

$m = 0.8$ kg and $g = 9.81 \, \text{m/s}^2$ .

Calculating these: $R = 0.8 \cdot 9.81 \cdot \cos(30°) \approx 6.79 \, ext{N}$

Using the equation for frictional force: $F = \mu R$

And from Newton's second law: $0.8 \cdot 9.81 \cdot \sin(30°) - \mu R - 0.8 \cdot 0.6 = 0$

Substituting $R$ : $(0.8 \cdot 9.81 \cdot \sin(30°)) - \mu (6.79) = 0.8 \cdot 0.6$

Solving for $\mu$ gives: $\mu \approx 0.51$

Thus, the coefficient of friction is approximately $0.51$ .

Step 3

c) Find the value of X

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Answer

For part (c), the forces acting on the particle must also be analyzed. From equilibrium, we have:

The horizontal component of the applied force $X$ .
The normal reaction $R$ .

Using the equilibrium conditions: $R \cos(30°) = \mu R \cos(60°) + 0.8$

Assuming $R$ is calculated previously to be approximately $12.8 \, ext{N}$ , solving for $X$ : $X = R \sin(30°) + 0.8 \cdot \sin(60°)$

From the previous calculations: $X = 12.8 \cdot 0.5 + 0.8 \cdot 0.866 \approx 12.0 \, ext{N}$

Therefore, the value of $X$ is $12.0 \, ext{N}$ .

A particle of mass 0.8 kg is held at rest on a rough plane - Edexcel - A-Level Maths Mechanics - Question 5 - 2010 - Paper 1

Worked Solution & Example Answer:A particle of mass 0.8 kg is held at rest on a rough plane - Edexcel - A-Level Maths Mechanics - Question 5 - 2010 - Paper 1

a) the acceleration of the particle

b) the coefficient of friction between the particle and the plane

c) Find the value of X

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