6 A lorry of mass 16 000 kg climbs a straight hill ABCD which makes an angle the
ID: 1353040 • Letter: 6
Question
6 A lorry of mass 16 000 kg climbs a straight hill ABCD which makes an angle theta with the horizontal, where sin theta = 1/20. For the motion from A to B, the work done by the driving force of the lorry is 1200kJ and the resistance to motion is constant and equal to 1240 N. The speed of the lorry is 15 ms^ - 1 at A and 12ms^ - 1 at B. (i) Find the distance AB. For the motion from B to D the gain in potential energy of the lorry is 2400 kJ. (ii) Find the distance BD. For the motion from B to D the driving force of the lorry is constant and equal to 7200 N. From B to C the resistance to motion is constant and equal to 1240 N and from C to D the resistance to motion is constant and equal to 1860 N. (iii) Given that the speed of the lorry at D is 7ms^ - 1, find the distance BC.Explanation / Answer
i) Let AB = x
Net workdone = W_lorry + W_resistance + W_gravity
= 1200*10^3 + 1240*x*cos(180) - m*g*sin(theta)*x
= 1.2*10^6 - 1240*x + 16000*9.8*(1/20)*x
= 1.2*10^6 - 1240*x - 7840*x
= 1.2*10^6 - 9080*x
We know, Net workdone = change in kinetic energy
1.2*10^6 - 9080*x = 0.5*m*(vB^2 - vA^2)
1.2*10^6 - 9080*x = 0.5*16000*(12^2 - 15^2)
1.2*10^6 - 9080*x = -0.648*10^6
9080*x = (1.2+0.648)*10^6
x = 1.848*10^6/9080
= 203.5 m <<<<<<<------------Answer
ii) gain in potentail energy = m*g*sin(theta)BD
2400*10^3 = 16000*9.8*(1/20)*BD
==> BD = 2400*10^3/(16000*9.8*(1/20))
= 306. 1 m <<<<<<<------------Answer
iii) Net workdone from B to D,
Wnet = 7200*306.1 - (BC*1240 + CD*1860) - 2400*10^3
0.5*16000*(7^2 - 12^2) = 7200*306.1 - (BC*1240 + CD*1860) - 2400*10^3
-7.6*10^5 = 2203920 - (BC*1240 + CD*1860) - 2.4*10^6
BC*1240 + CD*1860 = (0.76 - 2.4)*10^6 + 2203920
BC*1240 + CD*1860 = 563920 ----(1)
and BC + CD = 306.1 ---(2)
on solving equations 1 and 2, we get
we get, BC = 8.75 m
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