The engine of a train has a mass of 5x10^4 kg. it can accelerate six railway cars having a total mass of 3x10^5 kg by 0.4 m/s^2 (the 6 railway cars have a combined mass of 3x10^5). what acceleration can the same engine using the same force give to four railway cars having a total mass of 2x10^5kg?
it is proportional inversely
a1/a2=M2/M1
a1/a2=(3E5+5E4)/(2E5+5E4)
solve for a1 in terms of the original a2 acceleration.
let us examine it:
let F be the pulling force, mu the coefficent of friction.
Net pulling force= totalmass*a
F-mu(totalmass)=totalmass*a
F= totalmass(a+mu)
but F the pulling force remains the same on both trains
so
totalmass1(a1+mu)=totalmass2(a2+mu)
TM1*a1+ TM1*mu=TM2*a2+TM2*mu
a2= (TM1*a1+TM1*mu-TM2*mu)/TM2
in case one, TM1= 5E4+3E5, and in case two, TM2=(5E4+2E5)
Your teacher did not specific the friction. If you ignore friction, mu=0, but mu is hardly ever even close to zero on a freight train.
To find the acceleration that the engine can give to four railway cars, we need to apply Newton's second law of motion. According to the second law, force (F) is equal to the mass (m) multiplied by the acceleration (a), or F = m*a.
In the given scenario, we know that the engine can accelerate the six railway cars by 0.4 m/s^2. The total mass of the six railway cars is 3x10^5 kg. Therefore, we can calculate the force applied by the engine as follows:
F = m * a
F = (6x10^5 kg) * (0.4 m/s^2)
F = 2.4x10^5 N
Now, we can use the force calculated above to find the acceleration that the same engine can give to four railway cars. Again, using Newton's second law, we have:
F = m * a
a = F / m
a = (2.4x10^5 N) / (2x10^5 kg)
a = 1.2 m/s^2
Therefore, the same engine, using the same force, can give an acceleration of 1.2 m/s^2 to four railway cars with a total mass of 2x10^5 kg.