Q BgQuestion:

      
Rookie
Karma Points: 0
Respect (0%):
posted by  christinahughes80 on 11/3/2009 4:04:47 PM  |  status: Live  |  Earned Karma: 0

help help help homeworks due in 2 hours!! :(

Course Textbook Chapter Problem Needs by
N/A N/A N/A N/A 11/3/2009 at 5:00:00 PM
Question Details:
Okay here goes:
6) A 76 kg box slides down a 24 degree ramp with an acceleration of 3.2 m/s^2. Find the coefficent of kinetic friction between the box and the ramp.
7) What acceleration would a 161 kg mass have down this ramp?
8) A box of books weighing 346 N moves with a constant velocity across the floor when it is pushed witha  force of 417 N exerted downward at an angle of 31.9 degrees below the horizontal. Find the coefficent of kinetic friction between the box and the floor.
9) A freight train has a mass of 2.3 x 10^7 kg. If the locomotive can exert a constant pull of 7.5 x 10^5 N, how long would it take to increase the speed of the train from rest to 84.9 km/h? (Disregard friction.)
10) A 3.1 kg block is pushed along the ceiling with a constant applied force of 87 N that acts as an angle of 59 degrees with the horizontal. The block accelerates to the right at 6.4 m/s^2. What is the coefficent of kinetic friction between the block and the ceiling?
11) A 4.7 kg bag of groceries is in equilibrium on an incline of 30 degrees. What is the magnitude of the normal force on the bag?
All problems use 9.81 for gravity.
Someone please help me I CANT DO IT AAAHHHH!!!
Tags: Physics

AAnswers:

Answer Question Ask for clarification
Scholar
Karma Points: 298
(Calvin)
posted by eab35 on 11/3/2009 5:55:12 PM  |  status: Live
Asker's Rating: This answer has not been rated. If you asked this question, then please login.   
Response Details:
6)
Fg - Ff = F
mgsinθ - μmgcosθ = ma
μ = (gsinθ - a)/gcosθ = ((9.8)(sin24) - 3.2)/(9.8)(cos24) = .088
7)
mgsinθ - μmgcosθ = ma
mass cancels out of the equation so it would still have 3.2 m/s/s down the ramp
8)
Fnormal = 346 N + 417cos31.9 N = 700.02 N
Fx = 417sin31.9 N = 220.36 N
μFnormal = Fx
μ = Fx/Fnormal = (220.36)/700.02 = .315
9)
ma = F
a = F/m
a = (Δv)/Δt
Δv/t = F/m
t = Δvm/F
84.9 km/h = 23.58 m/s
t = 23.58(2.3x107)/(7.5x105) = 723.12 seconds = 12.05 minutes
10)
Do the same thing as in 8) with the new numbers
11)
Fnormal = mgcosθ = 4.7(9.8)cos30 = 39.9 N
(I used 9.8 instead of 9.81. I didnt see that until the end but you get the idea)
Answer Question Ask for clarificarion

Join Cramster's Community

Cramster.com brings together students, educators and subject enthusiasts in an online study community. With around-the-clock expert help and a community of over 100,000 knowledgeable members, you can find the help you need, whenever you need it. Join for free today » How Cramster is different from tutoring »