Question #131716

A wagon of mass 50 tones, starts from rest and travels 30 meters down a 2% grade and strikes a post with bumper spring as shown in Figure below. If the rolling resistance of the track is 60 N/t, find the velocity with which the wagon strikes the post. Also find the amount by which the spring will be compressed, if the bumper spring compresses 2 mm per 30 kN force.

Expert's answer

Weight of wagon(W)=50x1000x9.81 N

=490500N

Total Rolling resistance=60x50 N

=3000N

Acceleration of rolling resistance

(a_{R})=3000/mass=3000/(50x1000)

=0.06 m/s^{2}

Initial velocity (u)=0 m/s

Distance travelled along inclined path(d)=30m

Gradient of path(dy/dx)=-2%=-0.20

If is the angle of inclination of path with horizontal.

Tan =dy/dx=0.02

= 1.146 degrees

Acceleration due to gravity acts in the vertical direction (g)=9.81m/s^{2}

Acceleration due to gravity along the inclined path=g x sin =9.81xsin(1.146)

=0.196 m/s^{2}

Net acceleration along the path

(a)=0.196-a_{R}=0.136 m/s^{2}

By kinematical equation

V^{2}=u^{2}+2 x a x d

V=square root of(2x 0.136x30)

V=2.856m/s

If X is the compression in the spring of spring constant K.

Potential energy of spring (PE)= Kinetic energy of wagon (KE)

(1/2).K.X^{2}=(1/2).m.V^{2}

(30000/0.002).X^{2}=(50000).(2.856)^{2}

Compression in the spring (X)

=0.02718 m=27.18 mm

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