a 3000kg trunk moving at a speed of stop after covering some distance the force applical by breaks is 27000N computs the distances covered and work done by this force
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YOUR QUESTION IS INCOMPETE
SO LET THE INITIAL SPEED BE 90 m/s
You have forget to 90 after speed
Mass of the truck = 3000kg
Force applied = 27000 N
As we know that F = ma
27000 = 3000×a
a = 9 m/s²
As the truck has stopped so final velocity will be zero, so v=0 and acceleration will be negative a = -9 m/s²
v² - u² = 2as
-(90)² = 2(-9)s
8100 = 18s
s= 450m
So truck stops after 450m
Work done = Fs
= 27000(450)
= 12150000 J
Work done by brakes will be -12150000 J. It is negative because it stopped the truck.
SO LET THE INITIAL SPEED BE 90 m/s
You have forget to 90 after speed
Mass of the truck = 3000kg
Force applied = 27000 N
As we know that F = ma
27000 = 3000×a
a = 9 m/s²
As the truck has stopped so final velocity will be zero, so v=0 and acceleration will be negative a = -9 m/s²
v² - u² = 2as
-(90)² = 2(-9)s
8100 = 18s
s= 450m
So truck stops after 450m
Work done = Fs
= 27000(450)
= 12150000 J
Work done by brakes will be -12150000 J. It is negative because it stopped the truck.
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