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Nov 2002 p4 q6
3591
(i) A particle P of mass 1.2 kg is released from rest at the top of a slope and starts to move. The slope has length 4 m and is inclined at 25° to the horizontal. The coefficient of friction between P and the slope is \(\frac{1}{4}\). Find
the frictional component of the contact force on P,
the acceleration of P,
the speed with which P reaches the bottom of the slope.
(ii) After reaching the bottom of the slope, P moves freely under gravity and subsequently hits a horizontal floor which is 3 m below the bottom of the slope.
Find the loss in gravitational potential energy of P during its motion from the bottom of the slope until it hits the floor.
Find the speed with which P hits the floor.
Solution
(i) (a) The frictional force \(F\) is given by \(F = \mu mg \cos \theta\). Substituting the values, \(F = 0.25 \times 1.2 \times 9.8 \times \cos 25^{\circ} = 2.72 \text{ N}\).
(b) Using Newton's second law, \(1.2g \sin 25^{\circ} - 2.719 = 1.2a\). Solving for \(a\), \(a = 1.96 \text{ m/s}^2\).