The railway of a train is shown in Fig. 6-41 (h1 = 37 m, h2 = 15 m, h3 = 20 m).

Q: The railway of a train is shown in Fig. 6-41 (h1 = 37 m, h2 = 15 m, h3 = 20 m).

Let the slope be an inclined plane approximately, a= angel of inclination, F=mgsin(a)-frictional force; F=mgsin(a)-0.14*mg; ma=mgsin(a)-0.14mg; a=g*sin(a)-0.14*g; sin (a) = h1/(distance travelled) =37/55; =>a=g(37/55)-0.14*g; =>a=0.537*g; =>a=5.22 m/s^2; (v^2)-(u^2)=2*a*s; (v^2)=(2.2^2)+(2*5.22*55); (v^2)=579.04; v=24.063 m/s.

ID: 1285482 • Letter: T

Question

The railway of a train is shown in Fig. 6-41 (h1 = 37 m, h2 = 15 m, h3 = 20 m). The train passes point 1 with a speed of 2.20 m/s. If the average force of friction is equal to 14% of its weight, with what speed will the train reach point 2? The distance traveled is 55.0 m.

Explanation / Answer

Let the slope be an inclined plane approximately,

a= angel of inclination,

F=mgsin(a)-frictional force;

F=mgsin(a)-0.14*mg;

ma=mgsin(a)-0.14mg;

a=g*sin(a)-0.14*g;

sin (a) = h1/(distance travelled) =37/55;

=>a=g(37/55)-0.14*g;

=>a=0.537*g;

=>a=5.22 m/s^2;

(v^2)-(u^2)=2*a*s;

(v^2)=(2.2^2)+(2*5.22*55);

(v^2)=579.04;

v=24.063 m/s.

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The railway of a train is shown in Fig. 6-41 (h1 = 37 m, h2 = 15 m, h3 = 20 m).

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