Question
A railway truck on a level, straight track is initially at rest. The truck is given a quick, horizontal push by an engine so that it now rolls along the track.
push from engine
railway truck
v = 4.3 m s–1
The engine is in contact with the truck for a time T = 0.54 s and the initial speed of the truckafterthepushis4.3ms–1. Themassofthetruckis2.2×103kg.
Due to the push, a force of magnitude F is exerted by the engine on the truck. The sketch shows how F varies with contact time t.
(i) Determine the magnitude of the maximum force Fmax exerted by the engine on
the truck. [4]
.................................................................. .................................................................. .................................................................. .................................................................. .................................................................. .................................................................. .................................................................. ..................................................................
pls help
push from engine
railway truck
v = 4.3 m s–1
The engine is in contact with the truck for a time T = 0.54 s and the initial speed of the truckafterthepushis4.3ms–1. Themassofthetruckis2.2×103kg.
Due to the push, a force of magnitude F is exerted by the engine on the truck. The sketch shows how F varies with contact time t.
(i) Determine the magnitude of the maximum force Fmax exerted by the engine on
the truck. [4]
.................................................................. .................................................................. .................................................................. .................................................................. .................................................................. .................................................................. .................................................................. ..................................................................
pls help
Answers
bobpursley
consider the tuck
F*time=impulse=mass*change invelocity
Force=masstruck*4.3/.54 N
Now that is the average force.
It goes from max, to zero. So Fmax=2*Favg
= 2*masstruck*4.3/.54
F*time=impulse=mass*change invelocity
Force=masstruck*4.3/.54 N
Now that is the average force.
It goes from max, to zero. So Fmax=2*Favg
= 2*masstruck*4.3/.54
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