Question

. A train is moving with a speed of 72 km/h on a curved
railway-line of 400 m radius. A spring balance loaded
with a block of weight of 5 kg is suspended from the roof of the trainwhat would be the reading of balance ?g=10.
ans=5.025​

Answer 1

Answer:

50.25

Explanation:

Given A train is moving with a speed of 72 km/h on a curved

railway-line of 400 m radius. A spring balance loaded

with a block of weight of 5 kg is suspended from the roof of the train what would be the reading of balance  

We know that  

Tension T in the spring of spring balance is given by

T cos θ = mg and T sin θ = mv^2 / r

By squaring and adding we get  

T^2 = (mg)^2 + (mv^2 / R)^2  

T = m √g^2 + v^4 / r^2

Now we know that v = 72 km / hr = 72,000 / 3600 = 20 m/s, r = 400 m, m = 5 kg g = 10 m / s^2

So T = 5 x √ 10^2 + 20^4 / 400^2

T = 5 x √ 100 + 160000 / 160000

T = 5 x √101

T = 50.25 N

Answer 2

Answer:

: M = 5.025 kg or T= 50.25 N

Explanation:

please have a look on the picture before going through this explanation.

5 kg block is subjected to two forces due to gravity, i.e. 50 N (F = mg)and centripetal force  

Calculating first the centripetal force;

F” = mv^2/R

M= 5 kg

R= 400m

(train speed 72 km/hr = 20 m/s )

V= 20m/s

Hence centripetal force is  

= [5 x (20)^2]/400

=5N

.

Now calculating the resultant force ;

= √((50)^2+(5)^2 )

T= 50.25

Now weight on balance;

F = m g;

M = 50.25/10

M = 5.025 kg