Force Problems 3 & 4 [SFHS99 4.P.22 & 23]

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The net force on the 1680 kg car moving at a constant velocity of 2.94 m/s is zero, as per Newton's first law of motion, which applies regardless of the direction of movement. For the freight train with a mass of 5.5 x 10^7 kg and a locomotive exerting a pull of 15 x 10^5 N, the acceleration can be calculated using F=ma, resulting in an acceleration of 2.75 x 10^-4 m/s². To reach a speed of 85 km/h (23.61 m/s) from rest, it would take approximately 85972 seconds, or about 23 hours and 52 minutes, under the given force. If the force were instead 15 x 10^5 N, the time required would be significantly reduced to 865.7 seconds. The calculations demonstrate the relationship between force, mass, and acceleration in determining motion.
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3. [SFHS99 4.P.22.] An 1680 kg car is moving to the right at a constant velocity of 2.94 m/s.

(a) What is the net force on the car?
wrong check mark N

(b) What would be the net force on the car if it were moving to the left?
in Newtons = ??




4. [SFHS99 4.P.23.] A freight train has a mass of 5.5 107 kg. If the locomotive can exert a constant pull of 15 105 N, how long would it take to increase the speed of the train from rest to 85 km/h?
in seconds =??

thanks alot
 
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For your first question.

The net force will be zero if there is no acceleration! Newtons first law of motion.

same would apply to part b if it moves at constant velocity towards left.

For second question.

we know the mass and we know the force it can apply.
F=ma

\frac{F}{m}=a

\frac{15105}{5.5*10^7}=a

a=2.75*10^-^4

v=u+at

velocity is 85KM/h which is 23.61 m/s

23.61=0+2.75*10^-^4t

t=\frac{23.61}{2.75*10^-^4}

t=85972 s

23 hours 52 mins 52 seconds!

if the force applied by engine is 15105N.

if it is 15*10^5 then the answer is 865.7 seconds
 
thanks a lot they are correct, your explanations helped alot
 
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