A particle P of mass 4 kg is moving up a fixed rough plane at a constant speed of 16 ms^-1 under the action of a force of magnitude 36 N - Edexcel - A-Level Maths Mechanics - Question 8 - 2012 - Paper 1

The coefficient of friction (µ) can be calculated using:

$F_{f} = \mu R$

From our derived values, we will use the equation:

$36 \cos 30° = F + 4g \sin 30°$

Rearranging this gives us:

$\mu = \frac{36 \cos 30° - 4g \sin 30°}{R}$

Substituting R into this equation allows us to compute µ. Once values are substituted:

1. $4g \sin(30°) \approx 19.62$ N
2. Inserting into the overall expression:

$\mu \approx \frac{36 \times \frac{\sqrt{3}}{2} - 19.62}{15.9} \approx 0.726$

When the force is removed, the particle will start decelerating due to the frictional force. The new normal reaction R becomes:

$R = 4g \cos 30°$

Then applying Newton's second law, we have:

$-4g \cos 30° - 4g \sin 30° = 4a$

Here, 'a' is the acceleration, and substituting the known values allows for further computation:

1. Calculate the new acceleration: $a = -11.06 \, m/s^2$
2. Using the equation of motion to find distance (s): $v^2 = u^2 + 2as$
3. Where the initial velocity u = 16 m/s and final velocity v = 0 m/s, rearranging gives: $s = \frac{-u^2}{2a}$
4. Plugging in the numbers yields: $s = \frac{-(16)^2}{2 \times (-11.06)} \approx 11.6 \, m$