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sally_hayes sally_hayes
wrote...
11 years ago
I've been having trouble with this question 59 in the textbook:

Sam, whose mass is 75kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 200 N and a coefficient of kinetic friction on snow of 0.10. Unfortunately, the skis run out of fuel after only 10s.
a. What is Sam's top speed?
b. How far has Sam traveled when he finally coasts to a stop?

The book says the answer to a. is 17m/s, which I can't figure out how they got. If anyone can figure it out and show how the answer comes to 17m/s, that'll be great. Thanks. (:
Source  College Physics: Second Edition
Read 20499 times
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wrote...
11 years ago
First you must find the acceleration of the jet ski before it ran out of fuel
this can be done using newton's second law that states that the sum of forces acting on an object equals the mass of that object times its acceleration . (sigma F = m . a) there are two forces acting on the jet ski before it runs out of fuel they are
the thrust force from the engine = 200 N
the frictional force which equals Friction = coeificant of kinetic friction times the normal

according to newton's third law the normal force exactly equals the weight of the jet ski (75 x 10 m/s^2   = 750 N) as a result the force of friction can be calculated to be F = 0.1 * 750 = 75 N

the sum of forces equal F (thrust) - F (friction) = m x a
200-75 = 75 x a
(we subtract them since the forces are in opposite direction)
the direction of friction is always opposite to motion
a=1.6666 m/s^2

know that we know the acceleration we can calculate the max speed (the speed after 10 seconds pass) since the motion is uniformly accelerated
V = a. t  = 1.6666 x 10 = 16.6666 m/s ~ 17 m/s

b) after the engine stops the acceleration is now different since the jet ski is only acted upon by the frictional force (thrust = 0 N)
similarly we calculate the acceleration F = m.a    -75 = 75 .a   and a =-1 m/s^2

since the motion is now uniformly decelerated we can calculate the distance traveled using the formula V(final)^2  - V(initial)^2  = 2 x a x d

V(final) = 0 since we want to know the distance traveled (d) when it stops
and V(initial) = 17 m/s as calculated previously and a =-1 m/s^2
- (17^2) = 2 x (-1) x d
then d = 144.5 m
sally_hayes Author
wrote...
11 years ago
First you must find the acceleration of the jet ski before it ran out of fuel
this can be done using newton's second law that states that the sum of forces acting on an object equals the mass of that object times its acceleration . (sigma F = m . a) there are two forces acting on the jet ski before it runs out of fuel they are
the thrust force from the engine = 200 N
the frictional force which equals Friction = coeificant of kinetic friction times the normal

according to newton's third law the normal force exactly equals the weight of the jet ski (75 x 10 m/s^2   = 750 N) as a result the force of friction can be calculated to be F = 0.1 * 750 = 75 N

the sum of forces equal F (thrust) - F (friction) = m x a
200-75 = 75 x a
(we subtract them since the forces are in opposite direction)
the direction of friction is always opposite to motion
a=1.6666 m/s^2

know that we know the acceleration we can calculate the max speed (the speed after 10 seconds pass) since the motion is uniformly accelerated
V = a. t  = 1.6666 x 10 = 16.6666 m/s ~ 17 m/s

b) after the engine stops the acceleration is now different since the jet ski is only acted upon by the frictional force (thrust = 0 N)
similarly we calculate the acceleration F = m.a    -75 = 75 .a   and a =-1 m/s^2

since the motion is now uniformly decelerated we can calculate the distance traveled using the formula V(final)^2  - V(initial)^2  = 2 x a x d

V(final) = 0 since we want to know the distance traveled (d) when it stops
and V(initial) = 17 m/s as calculated previously and a =-1 m/s^2
- (17^2) = 2 x (-1) x d
then d = 144.5 m

Thanks for the help on part a. I see what you did for part b, and it makes sense, but the back of the book says the answer for part b. is 230m (not sure how they got that). I probably should have stated that in the beginning but I thought I could figure it out from the answer to part a. Not sure if the answer in the book is a mistake, or...
wrote...
11 years ago Edited: 11 years ago, kh006
144.5 is the distance traveled from the moment the engine runs out of fuel till the jet ski completely stops , it seems they want the distance traveled from the begining , just add the add the distance traveled during the first 10 seconds

d = 0.5 * a * t^2
d= 0.5 * 1.666 * (10^2) = 83.3 m
total distance = 83.3 + 144.5 = 227.83 m ~ 230 m
sally_hayes Author
wrote...
11 years ago
144.5 is the distance traveled from the moment the engine runs out of fuel till the jet ski completely stops , it seems they want the distance traveled from the begining , just add the add the distance traveled during the first 10 seconds

d = 0.5 * a * t^2
d= 0.5 * 1.666 * (10^2) = 83.3 m
total distance = 83.3 + 144.5 = 227.83 m ~ 230 m

Okay. Thank you!! (:
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