Answer to Question #131716 in Mechanical Engineering for JONES APPAH

Question

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.

Answer 1

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/s²

Initial velocity (u)=0 m/s

Distance travelled along inclined path(d)=30m

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

If $\empty$ is the angle of inclination of path with horizontal.

Tan$\empty$ =dy/dx=0.02

$\empty$= 1.146 degrees

$\empty$

Acceleration due to gravity acts in the vertical direction (g)=9.81m/s²

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

=0.196 m/s²

Net acceleration along the path

(a)=0.196-a_R=0.136 m/s²

By kinematical equation

V²=u²+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.X²=(1/2).m.V²

(30000/0.002).X²=(50000).(2.856)²

Compression in the spring (X)

=0.02718 m=27.18 mm