A box of mass 2 kg is pulled up a rough plane face by means of a light rope. The plane is inclined at an angle of 20° to the horizontal, as shown in Figure 2. The rope is parallel to a line of greatest slope of the plane. The tension in the rope is 18 N. The coefficient of friction between the box and the plane is 0.6. By modelling the box as a particle, find (a) the normal reaction of the plane on the box, (b) the acceleration of the box.

A-Level Edexcel Maths Mechanics Forces: A box of mass 2 kg is pulled up

A box of mass 2 kg is pulled up a rough plane face by means of a light rope - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Question 4

A box of mass 2 kg is pulled up a rough plane face by means of a light rope. The plane is inclined at an angle of 20° to the horizontal, as shown in Figure 2. The ro... show full transcript

Worked Solution & Example Answer:A box of mass 2 kg is pulled up a rough plane face by means of a light rope - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Step 1

Find the normal reaction of the plane on the box

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Answer

To find the normal reaction (R) acting on the box, we consider the forces acting perpendicular to the inclined plane. The weight of the box can be resolved into two components: one perpendicular to the plane and one parallel to the plane.

The weight of the box is given by: $W = mg = 2 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 19.6 \text{ N}$

The perpendicular component of the weight is: $W_{\perp} = W \cos(20°) = 19.6 \cos(20°)$

Thus, the normal reaction R can be determined using the equation: $R = W_{\perp} + T \sin(20°)$

Substituting the tension (T = 18 N): $R = 19.6 \cos(20°) + 18 \sin(20°)$

Calculating this gives: $R \approx 18.4 \text{ N} \text{ (or } 18 \text{ N)}$ .

Step 2

Find the acceleration of the box

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Answer

To find the acceleration (a) of the box, we consider the forces acting parallel to the plane. The net force acting on the box in the direction of motion can be expressed as:

$F_{net} = T - W_{\parallel} - F$

where:

The parallel component of weight is given by: $W_{\parallel} = W \sin(20°) = 19.6 \sin(20°)$
The frictional force (F) is given by: $F = \mu R = 0.6 R$

Substituting these into the net force equation, we get: $F_{net} = T - W \sin(20°) - 0.6 R$

Using R from part (a), we then substitute into the equation: $18 - 19.6 \sin(20°) - 0.6(18.4) = 2a$

Solving for a gives: $a = \frac{18 - 19.6 \sin(20°) - 0.6(18.4)}{2}$

Calculating this yields: $a \approx 0.123 \text{ m/s}^2 \text{ (or } 0.12 \text{ m/s}^2)$ .

A box of mass 2 kg is pulled up a rough plane face by means of a light rope - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Worked Solution & Example Answer:A box of mass 2 kg is pulled up a rough plane face by means of a light rope - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Find the normal reaction of the plane on the box

Find the acceleration of the box

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