1. Jason takes of across level water on his jet-powered skis. The

combined mass of Jason and skis is 75 kg. The skis have a thrust of

200 N and a coeffcient of kinetic friction on water of 0.1. Unfortunately,

the skis run out of fuel after 57 s. What is Jason's top speed?

 

I am just doing sample problems for my test today please help

thank you

You have enough information to use the basic kinematics equation vf = vo+at. The force in the forward direction is equal and opposite to the thrust, 200N, minus the force due to friction (F-friction = umg = 0.1*75*9.81). Divide that force by mass to find acceleration, and plug it into the equation. vf is your top speed.

http://www.scienceforums.net/forum/showthread.php?t=35529&highlight=Skis

 

There is a fair amount of info here also.

Oky

I am trying to work this problem out myself

but I got stuck

I have sollution as well

I got till this far

 

 

1. F = FT - Mu kmg

F = 200 - (0:1)(75)(9:8) = 126 N

F = ma = F/m = 126=75 = 1:69 m/s2

s = 1/2at2 = 1/2(1:69)(61)2 = 3144 m<---------- I don't get how they got and where they got 61 from

v = at = (1:69)(61) = 103 m/s ]<---------- same for here

x = v2/2a; a = Mu kg

x = (103)2

2(0:1)(9:8) = 5413

xT = 3144 + 5413 = 8:6 km

You are over-analyzing the problem. All the new information you need is the acceleration, and you already know the time over which acceleration occurs.

Oky

I am trying to work this problem out myself

but I got stuck

I have sollution as well

I got till this far

 

 

1. F = FT - Mu kmg

F = 200 - (0:1)(75)(9:8) = 126 N

F = ma = F/m = 126=75 = 1:69 m/s2

s = 1/2at2 = 1/2(1:69)(61)2 = 3144 m<---------- I don't get how they got and where they got 61 from

v = at = (1:69)(61) = 103 m/s ]<---------- same for here

x = v2/2a; a = Mu kg

x = (103)2

2(0:1)(9:8) = 5413

xT = 3144 + 5413 = 8:6 km

 

Are you sure the solution is to the exact same problem? Sometimes similar problems are written with different values in them.