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

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SUMMARY

The discussion addresses two physics problems from SFHS99, specifically problems 4.P.22 and 4.P.23. For problem 4.P.22, the net force on a 1680 kg car moving at a constant velocity of 2.94 m/s is zero, as per Newton's first law of motion. In problem 4.P.23, a freight train with a mass of 5.5 x 107 kg, subjected to a pulling force of 15 x 105 N, would take approximately 85972 seconds to reach a speed of 85 km/h (23.61 m/s) if the force is 15 x 105 N, or 865.7 seconds if the force is 15105 N.

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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 a lot
 
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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.
[tex]F=ma[/tex]

[tex]\frac{F}{m}=a[/tex]

[tex]\frac{15105}{5.5*10^7}=a[/tex]

[tex]a=2.75*10^-^4[/tex]

[tex]v=u+at[/tex]

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

[tex]23.61=0+2.75*10^-^4t[/tex]

[tex]t=\frac{23.61}{2.75*10^-^4}[/tex]

[tex]t=85972 s[/tex]

23 hours 52 mins 52 seconds!

if the force applied by engine is 15105N.

if it is [tex]15*10^5[/tex] then the answer is 865.7 seconds
 
thanks a lot they are correct, your explanations helped a lot
 
Last edited:

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