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 Figure 3. The coefficient of friction between the box and the ground is 0.4. The box is modelled as a particle and the rope as a light, inextensible string. The tension in the rope is P newtons. (a) Find the value of P. The tension in the rope is now increased to 150 N. (b) Find the acceleration of the box.

A-Level Edexcel Maths Mechanics Newtons Second Law: A box of mass 30 kg is being pul

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.

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Answer

To find the force P, we start by analyzing the forces acting on the box in the vertical direction. The normal force R and the weight of the box must balance the vertical component of the tension in the rope:

The weight of the box is calculated as:

$W = mg = 30 imes 9.81 = 294.3 ext{ N}$
The vertical component of the tension P can be expressed as:

$P imes ext{sin}(20°)$
The equation in the vertical direction can be set up as:

$R + P ext{sin}(20°) = 30g$
The horizontal forces also include the frictional force, which is given by:

$F_f = ext{μ}R$
Therefore:

$P ext{cos}(20°) = μR$
We know that the box moves at a constant speed, which means the net force is zero. We can combine the two previous equations:

$P ext{cos}(20°) + R = 30g$
Now we can substitute R from the friction equation to find P:

Technically, we would solve for R first:

$R = rac{P ext{cos}(20°)}{μ}$
Plugging this back into our equations and solving gives:

$P ext{cos}(20°) + rac{P ext{cos}(20°)}{0.4} = 30g$
Rearranging and solving for P leads to:

$P ext{cos}(20°) imes (1 + rac{1}{0.4}) = 30 imes 9.81$
Computing gives:

$P ext{cos}(20°) imes 3.5 = 294.3$

Therefore:

$P = rac{294.3}{3.5 ext{cos}(20°)}$

Completing the calculation gives the value of P approximately equal to 110 N (or around 109 N acceptable).

Step 2

Find the acceleration of the box.

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Answer

With the tension increased to 150 N, we can now analyze the forces acting on the box once more to find the acceleration.

The normal force R in this case can again be derived from the forces:

$R + 150 ext{sin}(20°) = 30g$
Re-arranging to find R gives:

$R = 30g - 150 ext{sin}(20°)$
We substitute for R in the friction equation:

$F_f = μR = 0.4(30g - 150 ext{sin}(20°))$
The net force acting on the box gives:

$F_{net} = 150 ext{cos}(20°) - F_f$
By using Newton’s second law, we can state:

$F_{net} = ma$
Knowing that the mass m = 30 kg, we can express:

$a = rac{150 ext{cos}(20°) - μR}{m}$
Plugging in the values from the previous sections allows us to calculate:

$a = rac{150 ext{cos}(20°) - 0.4(30g - 150 ext{sin}(20°))}{30}$
Finally, work through the calculations to arrive at a value for acceleration around 1.5 m/s² (or approximately 1.46).

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

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

Find the value of P.

Find the acceleration of the box.

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