Question
A particle of mass 0.4 kg is projected with a speed of 12 m
A particle of mass 0.4 kg is projected with a speed of 12 \(ms^{-1} \) up a line of greatest slope of a smooth plane inclined at 30Å to the horizontal.
(i) Find the initial kinetic energy of the particle. up a line of greatest slope of a smooth plane inclined at 30Å to the horizontal.
(ii) Use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest.
Answer/Explanation
(i) KE = \(1⁄2 × 0.4 \times 12^{2}=28.8j\)
(ii) PE gain = 0.4gh [= 4d sin 30]
4h = 28.8
h = 7.2 h = d sin 30 d = 14.4 m
Question
A lorry of mass 14 000 kg moves along a road starting from rest at a point O. It reaches a point A, and then continues to a point B which it reaches with a speed of 24 m s−1. The part OA of the road is straight and horizontal and has length 400 m. The part AB of the road is straight and is inclined downwards at an angle of θ° to the horizontal and has length 300 m.
(i) For the motion from O to B, find the gain in kinetic energy of the lorry and express its loss in potential energy in terms of θ. [3]
The resistance to the motion of the lorry is 4800 N and the work done by the driving force of the lorry from O to B is 5000 kJ.
(ii) Find the value of 1. [3]
Answer/Explanation
(i) For KE gain = 4032 × 103 or PE loss = 42 × 106 sinθ
PE loss = 42 × 106 sinθ or KE gain = 4032 × 103
(ii) 5000 = 4032 – 42000sin θ + 3360
θ = 3.3°
Question
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 α to the horizontal, where sin α = 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 curveOA 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.
(i) 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
(ii) the acceleration of P when it is moving from A to B,
(iii) the speed of P when it reaches B.
Answer/Explanation
(i) PE loss = 2g(10 – 10 × 0.28)
[ 1⁄2\( 2y^{2}\)= 144]
= PE loss
Speed is 12\( ms^{–1}\)
(ii) R = 2g x 0.96
[2g × 0.28 – 2g × 0.96 ÷ 12 = 2a]
Acceleration is 2\( ms^{–1}\)
(iii) \(v^{2}\)= \(12^{2} \)+ 2 × 2 × 10]
Speed is\( 13.6 ms^{-1}\)