Exam-Style Problem

June 2012 p41 q5

3558

The diagram shows the vertical cross-section OAB of a slide. The straight line AB is tangential to the curve OA at A. The line AB is inclined at \(\alpha\) to the horizontal, where \(\sin \alpha = 0.28\). The point O is 10 m higher than B, and AB has length 10 m (see diagram). The part of the slide containing the curve OA is smooth and the part containing AB is rough. A particle P of mass 2 kg is released from rest at O and moves down the slide.

Find the speed of P when it passes through A.
The coefficient of friction between P and the part of the slide containing AB is \(\frac{1}{12}\). Find
1. the acceleration of P when it is moving from A to B

Solution

(i) The potential energy (PE) loss from O to A is given by:

\(\text{PE loss} = 2g(10 - 10 \times 0.28)\)

Using conservation of energy, \(\frac{1}{2} mv^2 = \text{PE loss}\):

\(\frac{1}{2} \times 2 \times v^2 = 144\)

Solving for \(v\), we find \(v = 12 \text{ m/s}\).

(ii) The normal reaction \(R\) is given by:

\(R = 2g \times 0.96\)

Using Newton's second law, the equation of motion is:

\(2g \times 0.28 - 2g \times 0.96 \times \frac{1}{12} = 2a\)

Solving for \(a\), we find \(a = 2 \text{ m/s}^2\).

(iii) Using the equation \(v^2 = u^2 + 2as\) from A to B:

\(v^2 = 12^2 + 2 \times 2 \times 10\)

Solving for \(v\), we find \(v = 13.6 \text{ m/s}\).

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