A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25° to the horizontal. The particle passes through two points A and B, where AB = 10 m, as shown in Figure 3. The speed of P at A is 2 m s⁻¹. The particle P takes 3.5 s to move from A to B. Find (a) the speed of P at B, (b) the acceleration of P, (c) the coefficient of friction between P and the plane.

A-Level Edexcel Maths Mechanics Constant Acceleration - 1D: A particle P of mass 0.6 kg slid

A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 5 - 2013 - Paper 1

Question 5

A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25° to the horizontal. The particl... show full transcript

Worked Solution & Example Answer:A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 5 - 2013 - Paper 1

Step 1

the speed of P at B

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Answer

First, we use the equation of motion for speed:

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

Here, we know:

Initial speed, $u = 2$ m/s
Distance, $s = 10$ m
Time, $t = 3.5$ s

Plugging in these values:

$10 = 2 \cdot 3.5 + \frac{1}{2} a \cdot (3.5)^2$

We can rearrange this to find acceleration ( $a$ ):

$a = \frac{2(10 - 7)}{(3.5)^2} = \frac{6}{12.25} \approx 0.490 \, \text{m/s}^2$

Now, substituting back to find speed at B:

Using the equation:

$v = u + at$

Substituting the values:

$v = 2 + 0.490 \cdot 3.5 = 2 + 1.715 \approx 3.71 \, \text{m/s}$

Step 2

the acceleration of P

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Answer

From the earlier calculations, we have already found that:

$a \approx 0.490 \, \text{m/s}^2$

Step 3

the coefficient of friction between P and the plane

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Answer

First, we calculate the normal reaction ( $R$ ):

$R = mg \cos(25°) = 0.6 \cdot 9.81 \cdot \cos(25°)$

The frictional force is given by:

$F_{friction} = \\mu R$

Resolving forces parallel to the incline gives:

$F_{net} = mg \sin(25°) - F_{friction}$

Substituting:

$0.6 \cdot 9.81 \cdot \sin(25°) - \mu (0.6 \cdot 9.81 \cdot \cos(25°)) = 0.6 \cdot 0.490$

Finally, solving for $\,\mu$ will yield:

$\mu \approx 0.41 \text{ or } 0.411$

A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 5 - 2013 - Paper 1

Worked Solution & Example Answer:A particle P of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 5 - 2013 - Paper 1

the speed of P at B

the acceleration of P

the coefficient of friction between P and the plane

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