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16. A freight train consists of two 8.00×105 -kg engines and 45 cars with average masses of 5.50×105 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00×10−2 m /s 2 if the force of friction is 7.50×105 N , assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the magnitude of the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

Respuesta :

Answer:

Part a)

[tex]F = 8.175 \times 10^5 N[/tex]

Part b)

[tex]F = 4.97 \times 10^4 N[/tex]

Explanation:

Part a)

As we know that total mass of the engine and the cars is given as

[tex]M = 8 \times 10^5 kg[/tex]

[tex]m = 5.50\times 10^5 kg[/tex]

[tex]M + m = 13.5 \times 10^5 kg[/tex]

now we know the acceleration as

[tex]a = 5.00 \times 10^{-2} m/s^2[/tex]

Force of friction is given as

[tex]F_f = 7.50 \times 10^5 N[/tex]

now we have

[tex]F - F_f = (M + m) a[/tex]

[tex]F = F_f + (M + m)a[/tex]

[tex]F = (7.50 \times 10^5) + (13.5 \times 10^5)(5 \times 10^{-2})[/tex]

[tex]F = 8.175 \times 10^5 N[/tex]

Part b)

As we know that 37th car will exert force on three cars behind it

so the mass of three cars is

[tex]m = \frac{3}{45}(5.50\times 10^5)[/tex]

[tex]m = 3.67 \times 10^4 kg[/tex]

also for friction force we have

[tex]F_f = \frac{3}{47}(7.50 \times 10^5)[/tex]

[tex]F_f = 4.79 \times 10^4 N[/tex]

now we have

[tex]F - F_f = ma[/tex]

[tex]F = 4.79 \times 10^4 + (3.67 \times 10^4)(5 \times 10^{-2})[/tex]

[tex]F = 4.97 \times 10^4 N[/tex]

oladayoademosu

Answer:

a. -2.84 × 10⁵ N b. 104.46 × 10⁵ N

Explanation:

a. The net force on the freight train, ma = F - f where F = frictional force and f = backward force of engines. So, f = F - ma. Now, the total mass, m carried by the engine force = total mass of 45 trains cars + mass of two engines = 45 × 5.50 × 10⁵ + 2 × 8.00 × 10⁵ kg= (247.5 × 10⁵ + 16 × 10⁵) kg = 263.5 × 10⁵ kg.

Since the acceleration = 5.00 × 10⁻² m/s² and F = 7.50 × 10⁵ N.

So, f = F - ma = 7.50 × 10⁵ N - 263.5 × 10⁵ kg × 5.00 × 10⁻² m/s² = 7.50 × 10⁵ N - 13.175 × 10⁵N = -5.675 × 10⁵ N.

Since the engines exert identical forces, the force of each engine, f₀ = f/2 = -5.675 × 10⁵/2 N = -2.8375 × 10⁵ N ≅ -2.84 × 10⁵ N

b. After the 37 th cars, we have n - 37 cars. The net force on these cars is (n - 37)ma. The frictional force on these cars is F/(n - 37). Let f be the force on the couplings. So, the net force = frictional force on (n - 37) cars - force on couplings.

(n - 37)ma = F/(n - 37) - f

So, f = F/(n - 37) - (n - 37)ma. Since n = 45, we substitute the values of F, m and a

f = 7.50 × 10⁵/(45 - 37) N - (45 - 37)263.5 × 10⁵ kg × 5.00 × 10⁻² m/s²= 7.50/8 × 10⁵ N - 8 × 13.175 × 10⁵N = 0.9375 × 10⁵ N - 105.4 × 10⁵ N = -104.4625 × 10⁵ N. ≅ -104.46 × 10⁵ N.

So, the magnitude of the force between the 37th and 38th coupling is 104.46 × 10⁵ N

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