The diagram shows a motorbike of mass 300 kg being ridden along a straight road.
*insert picture of motorbike of mass 300 kg being ridden along a straight road*
The rider sees a traffic queue ahead. He applies the brakes and reduces the speed of the motorbike from 18 m/s to 3 m/s.
a) Use the equation in the box to calculate the kinetic energy lost by the motorbike.
Show clearly how you work out your answer.
kinetic energy = 1/2 × mass × speed^2
Kinetic energy lost = ........................................… J
(2 marks)
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The answer is 47250J. But how do you go about getting to this answer? I tried 18-13 = 15 and then doing 1/2 x 300 x 15^2 but it came out to something different. Help! I can't figure out how to do this :'(
Thank you!
*insert picture of motorbike of mass 300 kg being ridden along a straight road*
The rider sees a traffic queue ahead. He applies the brakes and reduces the speed of the motorbike from 18 m/s to 3 m/s.
a) Use the equation in the box to calculate the kinetic energy lost by the motorbike.
Show clearly how you work out your answer.
kinetic energy = 1/2 × mass × speed^2
Kinetic energy lost = ........................................… J
(2 marks)
-----------------------
The answer is 47250J. But how do you go about getting to this answer? I tried 18-13 = 15 and then doing 1/2 x 300 x 15^2 but it came out to something different. Help! I can't figure out how to do this :'(
Thank you!
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im doing the physics exam 2moro and its confusing as *****. but i luckily understand this question.
(0.5*300*18^2)-(0.5*300*3^2)=47250
bacially workout the kinetic energy for when he was going at 18m/s and 3m/s and get the difference.
could u check my physics question and see if u can answer it plz.
(0.5*300*18^2)-(0.5*300*3^2)=47250
bacially workout the kinetic energy for when he was going at 18m/s and 3m/s and get the difference.
could u check my physics question and see if u can answer it plz.
-
.5*300*18^2 - .5*300*3^2 .... you were half way there.. well kinda.