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 frictional force can be expressed as:

$F_f = \mu R$

Using the forces in the direction parallel to the inclined plane:

$36 \cos(30^\circ) = F + 4g \sin(30^\circ)$

Substituting values:

$36 \cdot \frac{\sqrt{3}}{2} = \mu \cdot 15.9 + 4 \cdot 9.81 \cdot \frac{1}{2}$

Solving this equation for ( \mu ):

$\mu = \frac{36 \cdot \frac{\sqrt{3}}{2} - 19.62}{15.9} = 0.726$

Thus, ( \mu \approx 0.73 ).

Once the force of magnitude 36 N is removed, the new normal reaction R can be expressed as:

$R = 4g \cos(30^\circ) - 4g \sin(30^\circ)$

Substituting values gives:

$R = 4 \cdot 9.81 \cdot \frac{\sqrt{3}}{2} - 4 \cdot 9.81 \cdot \frac{1}{2}$

Then we find the acceleration a:

$a = -\frac{4g \cos(30^\circ) - 4g \sin(30^\circ)}{4}$

Using ( v^2 = u^2 + 2as ) where ( u = 16,ms^{-1} ):

$16^2 = 0 + 2 \cdot a \cdot s$

Solving this gives:

$s = \frac{16^2}{2 \cdot 11.06} = 11.6\,m$

Thus, the distance that P travels is approximately 12 m.