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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
Question 6
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. The coefficient of friction betwee... show full transcript
Worked Solution & Example Answer: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
Step 1
(a) the frictional force acting on P as it moves up the plane
Answer
To determine the frictional force acting on particle P, we first calculate the normal reaction force (R) exerted on it. The weight of the particle (W) is given by:
The component of weight acting perpendicular to the inclined plane is:
R = W imes ext{cos}(30°) = 29.4 imes rac{ ext{√3}}{2} \approx 25.46 ext{ N}
The frictional force (F) is calculated using the coefficient of friction (μ = 0.4):
Thus, the frictional force acting on P as it moves up the plane is approximately 10 N.
Step 2
(b) the distance moved by P up the plane before P comes to instantaneous rest
Answer
To find the distance moved by P, we need to compute the acceleration (a) when it is projected up the slope. The net force acting on P when moving up is:
F_{ ext{net}} = -F - W imes ext{sin}(30°) = -10 - 3g imes rac{1}{2} = -10 - 14.7 \approx -24.7 ext{ N}
Using Newton's second law, we can find the acceleration:
To calculate the distance (s) before P comes to instantaneous rest, we use the equation of motion:
v^2 = u^2 + 2as\, ext{ where } u = 6 ext{ m/s, } v = 0\
Substituting the known values = 0, u = 6, and a = -8.23:
Thus, the distance moved by P up the plane before it comes to instantaneous rest is approximately 2.19 m.