Question

A train of mass 16,800 kg is pulled by its locomotive with a force of 10,000 N and is speed increases from 10 m/s to 12 m/s in a distance of 50 m. Calculate the resisting force assumed constant for that distance​

Answer 1

Explanation:

m=600

mℓ=125

mnet=725

f=

tan

10

t

net

=10×125

V=36km/h=36×

18

5

m/s

P=t

net

×V=72500W

Answer 2

Answer:

The resistive force is 7392N.

Explanation:

First, let's find the acceleration from the third equation of motion i.e.

$v^2-u^2=2as$         (1)

Where,

v=final velocity with which the body is moving

u=initial velocity with which the body is moving

a=acceleration of the body

s=distance traveled by the body

From the question we have,

Mass(M) of the train=16800kg

v=12m/s

u=10m/s

s=50m

By substituting the values in equation (1) we get;

$(12)^2-(10)^2=2a\times 50$

$144-100=100a$

$a=\frac{44}{100}=0.44m/s^2$                      (2)

The resisting force will be given as,

$F=Ma$                 (3)

By substituting the required values in equation (3) we get;

$F=16800\times 0.44=7392N$               (4)

Hence, the resistive force assumed constant is 7392N.