A particle P of mass 3 kg is projected up a line of greatest slope of a rough plane inclined at an angle of 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 6 - 2003 - Paper 1

To find the distance moved by particle P up the plane, we first need to calculate the deceleration (a) using Newton's second law. The forces acting on P while moving up the incline are:

• Gravitational force component down the incline:
  F_g = mg imes ext{sin}(30°) = 3 ext{ kg} imes 9.81 ext{ m/s}^2 imes rac{1}{2} = 14.715 ext{ N}

Using Newton's second law, we can set up the equation:

$F_{net} = F_{up} - (F_g + F_f)$
Where:

• $F_{up} = ma$ (Force due to particle's motion upwards)
• $F_g$ is the gravitational force component down the incline
• $F_f$ is the frictional force calculated earlier:

Combining these, we have:

$3a = 6 ext{ N} - (14.715 ext{ N} + 10.18 ext{ N})$

This leads to:

$3a = 6 - 24.895$

$3a = -18.895 ext{ N}$

Thus,

$a = -6.298 ext{ m/s}^2 ext{ (approximately 8.3 m/s}^2 ext{, acting downwards)}$

Now, we can use the kinematic equation to find the distance (s):

$v^2 = u^2 + 2as$

Where:

• $v = 0$ (final velocity, at rest)
• $u = 6 ext{ m/s}$ (initial velocity)
• $a = -8.3 ext{ m/s}^2$

Rearranging gives:

$0 = (6)^2 + 2(-8.3)s$

This simplifies to:

$0 = 36 - 16.6s$

Thus:

ightarrow ext{} s = rac{36}{16.6} ext{ m} ext{ (approximately 2.17 m)}$$ Therefore, the distance moved by P up the plane before it comes to instantaneous rest is approximately **2.17 m**.