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A particle P of mass 4 kg is moving up a fixed rough plane at a constant speed of 16 m s⁻¹ under the action of a force of magnitude 36 N - Edexcel - A-Level Maths Mechanics - Question 8 - 2012 - Paper 1
Question 8
A particle P of mass 4 kg is moving up a fixed rough plane at a constant speed of 16 m s⁻¹ under the action of a force of magnitude 36 N. The plane is inclined at 30... show full transcript
Worked Solution & Example Answer:A particle P of mass 4 kg is moving up a fixed rough plane at a constant speed of 16 m s⁻¹ under the action of a force of magnitude 36 N - Edexcel - A-Level Maths Mechanics - Question 8 - 2012 - Paper 1
Step 1
a) the magnitude of the normal reaction between P and the plane
Answer
To find the normal reaction force (R), we analyze the forces acting perpendicular to the inclined plane. The vertical force component is given by the sum of the weight of the particle and the vertical component of the applied force. The weight of the particle is calculated as:
The vertical component of the applied force is:
F_{vertical} = 36 imes rac{1}{2} = 18 ext{ N}
Using the angle of the incline, we can establish:
1}{2}$$ Substituting the known values: $$R + 18 = 39.24 imes rac{1}{2}$$ This simplifies to: $$R = 19.62 - 18 = 1.62 ext{ N}$$ Thus, the magnitude of the normal reaction R is approximately 15.9 N.Step 2
b) the value of μ
Answer
The coefficient of friction μ can be calculated using the equation:
We know that the horizontal forces are balanced at constant speed, which gives:
1}{2} = F + 4g imes rac{1}{2}$$ Substituting R into the equations yields: $$ rac{36 imes rac{1}{2}}{R} = μ$$ Rearranging gives: $$μ = rac{36 imes rac{1}{2} - 4g imes rac{1}{2}}{R}$$ Substituting R from part (a), we calculate: $$μ = rac{18 - 19.62}{15.9} = 0.726$$ Thus, the coefficient of friction μ is approximately 0.73.Step 3
c) Find the distance that P travels between the instant when the force is removed and the instant when it comes to rest
Answer
After the removal of the 36 N force, the net force acting on P can be given by:
R = 4g imes rac{1}{2}
Using the equations of motion, the initial velocity (u) is 16 m/s. Therefore, setting:
Where v = final velocity (0), u = 16 m/s, and a = -11.06 m/s² (deceleration due to gravity and friction).
So substituting values gives:
Rearranging leads to:
s = rac{16^2}{2 imes 11.06} = 11.6 ext{ m}
Thus, the distance that P travels before coming to rest is approximately 11.6 m.