Question

A freight train consists of two 8.00×104-kg engines and 45 cars with average masses of 5.50×104 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00×10–2m/s2 if the force of friction is 7.50×105N, assuming the engines exert identical forces?

Answer 1

Answer:

$4.41\cdot 10^5 N$

Explanation:

First of all, let's calculate the total mass of the train+the engines:

$m=2(8.00\cdot 10^4 kg) + 45(5.50\cdot 10^4 kg) = 2.64\cdot 10^6 kg$

Then we can apply Newton's second law, which states that the resultant of the forces is equal to the product between mass (m) and acceleration (a):

$\sum F = ma$ (1)

In this case there are two forces:

- The pushing force exerted by the engines, F

- The frictional force, $F_f = 7.50 \cdot 10^5 N$, in an opposite direction to the acceleration

So (1) becomes

$F-F_f = ma$

Since the acceleration must be

$a=5.00\cdot 10^{-2} m/s^2$

We  can solve the formula to find F:

$F=ma+F_f = (2.64\cdot 10^6 kg)(5.00\cdot 10^{-2} m/s^2) + 7.50 \cdot 10^5 N = 8.82\cdot 10^5 N$

However, this is the force exerted by both engines. So the force exerted by each engine must be half this value:

$F=\frac{8.82\cdot 10^5 N}{2}=4.41\cdot 10^5 N$