Question #247579
A wagon of mass 150 tonnes, starts from rest and travels 50 metres down a
5% grade and strikes a post with bumper spring as shown in Figure below.If the rolling resistance of the track is 50 N/t, find
a)Thevelocity with which the wagon strikes the post.
b)Also find the amount by which the spring will be compressed, if the bumper spring
compresses 1 mm per 25kNforce.
1
Expert's answer
2021-10-10T09:32:51-0400

a)

Using F=ma

1471500Sin2.86-50*150=150*1000*a

a=0.439m/s2

Now using v2=u2+2as

v2=0+2*0.439*50

v=6.625m/s


b)

E1=12mv2+mgh1\frac{1}{2}mv^{2}+mgh_1

E2=mgh2E_2=mgh_2

E3=12kx2E_3=\frac{1}{2}kx^{2}

xx is the compression of spring

E4=E_4= rolling resistance * xx

E1=E2+E3+E4E_1=E_2+E_3+E_4

12mv2+mgh1=mgh2+12kx2+50150x\frac{1}{2}mv^{2}+mgh_1=mgh_2+\frac{1}{2}kx^{2}+50*150*x

12mv2=mg(h2h1)+12kx2+50150x\frac{1}{2}mv^{2}=mg(h_2-h_1)+\frac{1}{2}kx^{2}+50*150*x

12mv2=mgSinθ+12kx2+50150x\frac{1}{2}mv^{2}=mg*Sin\theta +\frac{1}{2}kx^{2}+50*150*x

12150000(6.625)2=1500009.81Sin2.86+0.5\frac{1}{2}*150000*(6.625)^{2}=150000*9.81*Sin2.86+0.5

3292796.875=73421.50x+1225000000x2+7500x3292796.875=73421.50x+\frac{1}{2}*25000000x^{2}+7500x

391796.875=73421.5x+12500000x2+7500x391796.875=73421.5x+12500000x^{2}+7500x

12500000xx2+80921.5x3291796.875=012500000xx^{2}+80921.5x-3291796.875=0

x=0.50m


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