Question

A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How far must the student be pushed, starting from rest, so that her final kinetic energy is 352 J?

Answer 1

we have F=ma, i.e. a=F/m
according to info given in question a=45/m.....(1)
again,
(1/2)mv^2 = 352
or, v^2= 352 * 2/m.....(2)
Also we have v^2=2as.... (3) (starting from rest)
solving (1), (2) and (3) we get
s=7.82 m

hence the student must be pushed 7.82 m..

Answer 2

What does "a" stand for?

Answer 3

acceleration

Answer 4

alright Thank You! :D

Answer 5

you are most welcome.. :)

Answer 6

Another method:
By using principal of conservation of energy:
workdone=change in kinetic energy
F*d=(K.E)final-0;
d=(K.E)final/F=352/45=7.82m

Answer 7

Thank You!