Mechanics help
An engine and train weigh 250 tonnes. The engine exerts a pull of 150000. The resistance of motion is 1/100 of the weight of the train and it's braking force is 1/10 of the weight.
The train starts from rest and accelerates uniformly until it reaches a speed of 45 km/h. At this point the brakes are applied until the train stops.
Find the time taken for the train to stop to the nearest second.
Answer = 36.4
What I did :
F=ma
The resistance of motion=
1/100 × 250 000
= 2500
Braking force=
1/10 × 250 000
= 25000
Resultant force when car decelerates =
(2500 +25000)-150000
=-122500
F=ma
-122500=250000 a
a=-0.49
45 km/h = 12.5 m/s
t= (v-u) /a
= - 12.5/-0.49
=25.5
What did I do wrong?
The train starts from rest and accelerates uniformly until it reaches a speed of 45 km/h. At this point the brakes are applied until the train stops.
Find the time taken for the train to stop to the nearest second.
Answer = 36.4
What I did :
F=ma
The resistance of motion=
1/100 × 250 000
= 2500
Braking force=
1/10 × 250 000
= 25000
Resultant force when car decelerates =
(2500 +25000)-150000
=-122500
F=ma
-122500=250000 a
a=-0.49
45 km/h = 12.5 m/s
t= (v-u) /a
= - 12.5/-0.49
=25.5
What did I do wrong?
(edited 3 months ago)
Original post by raghadgamil
An engine and train weigh 250 tonnes. The engine exerts a pull of 150000. The resistance of motion is 1/100 of the weight of the train and it's braking force is 1/10 of the weight.
The train starts from rest and accelerates uniformly until it reaches a speed of 45 km/h. At this point the brakes are applied until the train stops.
Find the time taken for the train to stop to the nearest second.
Answer = 36.4
What I did :
F=ma
The resistance of motion=
1/100 × 250 000
= 2500
Braking force=
1/10 × 250 000
= 25000
Resultant force when car decelerates =
(2500 +25000)-150000
=-122500
F=ma
-122500=250000 a
a=-0.49
45 km/h = 12.5 m/s
t= (v-u) /a
= - 12.5/-0.49
=25.5
What did I do wrong?
The train starts from rest and accelerates uniformly until it reaches a speed of 45 km/h. At this point the brakes are applied until the train stops.
Find the time taken for the train to stop to the nearest second.
Answer = 36.4
What I did :
F=ma
The resistance of motion=
1/100 × 250 000
= 2500
Braking force=
1/10 × 250 000
= 25000
Resultant force when car decelerates =
(2500 +25000)-150000
=-122500
F=ma
-122500=250000 a
a=-0.49
45 km/h = 12.5 m/s
t= (v-u) /a
= - 12.5/-0.49
=25.5
What did I do wrong?
A few things.
•
You need to calculate the time for both the acceleration and the braking phases.
•
The mass is 250 000 kg and the weight is 250 000g N.
•
In the acceleration phase you have driving and resistance forces
•
In the braking phase you have braking and resistance forces.
Original post by mqb2766
A few things.
•
You need to calculate the time for both the acceleration and the braking phases.
•
The mass is 250 000 kg and the weight is 250 000g N.
•
In the acceleration phase you have driving and resistance forces
•
In the braking phase you have braking and resistance forces.
oh okay ill try again.
When accelerating :
F=ma
150000-25000 = 250 000 a
a = 0.5
t = 12.5/0.5
=25
When decelerating :
F=ma
250 000 +25000 = 250 000 a
a = -1.1
t= -12.5/-1.1
=11.36
total time = 36.4
Thanks! ( :
(edited 3 months ago)
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