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A box of mass 30 kg is being pulled along rough horizontal ground at a constant speed using a rope - Edexcel - A-Level Maths Mechanics - Question 6 - 2007 - Paper 1
Question 6
A box of mass 30 kg is being pulled along rough horizontal ground at a constant speed using a rope. The rope makes an angle of 20° with the ground, as shown in Figur... show full transcript
Worked Solution & Example Answer:A box of mass 30 kg is being pulled along rough horizontal ground at a constant speed using a rope - Edexcel - A-Level Maths Mechanics - Question 6 - 2007 - Paper 1
Step 1
Find the value of P.
Answer
To find the value of P, we start by considering the forces acting on the box. The normal force, R, and the force due to friction, are affected by the angle of the rope.
-
Using the frictional force equation:
where (\mu = 0.4). -
The vertical components of the forces give us:
( R + P \sin 20° = 30g )
where ( g \approx 9.81 ) m/s², leading to( R = 30g - P \sin 20° ) -
The horizontal components give us:
( P \cos 20° = \mu R )
Substituting R into the equation:
( P \cos 20° = 0.4(30g - P \sin 20°) ) -
Rearranging gives us:
[ P \cos 20° + 0.4 P \sin 20° = 0.4 \cdot 30g ]
[ P (\cos 20° + 0.4 \sin 20°) = 0.4 \cdot 30g ] -
Thus,
[ P = \frac{0.4 \cdot 30g}{\cos 20° + 0.4 \sin 20°} ]
Substituting ( g = 9.81 ):
[ P \approx \frac{0.4 \cdot 30 \cdot 9.81}{\cos 20° + 0.4 \sin 20°} \approx 109 \text{ N} \text{ (accept 110 N)} ]
Step 2
Find the acceleration of the box.
Answer
Given that the tension in the rope is now 150 N, we need to again analyze the forces acting on the box.
-
From the vertical forces, we have: ( R + 150 \cos 20° = 30g )
which gives:
( R = 30g - 150 \cos 20° \approx 30 imes 9.81 - 150 \times \cos 20° \approx 242.7 ) N. -
Using Newton's second law in the horizontal direction:
[ F_{net} = ma
ightarrow 150 \cos 20° - \mu R = 30a ]
Substituting for R:
[ 150 \cos 20° - 0.4 \cdot 242.7 = 30a ]
- Simplifying gives:
[ a = \frac{150 \cos 20° - 0.4 \cdot 242.7}{30} \approx 1.5 \text{ m/s}^2 \text{ (accept 1.46)}]