Open-Response Homework Problem 18.3 A 1-kg book requires a 0.5 N force to start
ID: 1444915 • Letter: O
Question
Open-Response Homework Problem 18.3 A 1-kg book requires a 0.5 N force to start it sliding across a table. Once it begins sliding, it requires a force of 0.25 N to maintain constant velocity. (a)[6 pts] What is the coefficient of static friction between the book and the table? (b)[5 pts] What is the coefficient of kinetic friction between the book and the table? Open-Response Homework Problem 18.4 When she was an undergrad, Liz used to work for a small company which produced and packaged sauces and other food items. She would routinely move 40kg crates of ingredients across the floor of the warehouse by pushing them at a velocity of 2m. These wooden crates had a coefficient s of friction of k = 0.1 with the polished concrete floor, and she would push with an applied force 30 from the horizontal. (a)[8 pts] Find the applied force. (b)[2 pts] Find the amount of work Liz had done after she had pushed one of these crates 10m across the warehouse floor. Open-Response Homework Problem 18.5 A block of inertia m is placed on a rough plane (coefficient of kinetic friction k) that is inclined by an angle to the horizontal. It is pressed against a spring of spring constant k parallel to the inclined surface until the spring is compressed from its relaxed state a distance d. It is then released and projected up the surface. (a)[4 pts] Draw an energy bar diagram for the energy while the spring is compressed before the object is released. Consider the earth to be part of your system. (b)[6 pts] Draw an energy bar diagram for the energy at the top of motion (with Earth still in the system). (c)[10 pts] In terms of the given quantities, how far will it travel on the incline before it has zero speed, as measured from the point of maximum compression of the spring? Call this distance x.
Explanation / Answer
18.3
A. a. 0.5 = mus*mg = mus*1*9.8
mus = 0.051
b. 0.25 = muk*mg
muk = 0.025
18.4
a. applied force = F
Horizontal force balance => Fcos(30) = mu*N
Vertical Force Balance => mg + Fsin(30) = N
so Fcos(30) = mu[mg+Fsin(30)]
0.866 F = 0.1[40*9.8 + F/2] = 39.2 + 0.05F
F = 48.039 N
b. Work done = Fcos(30)*10 = 416.031 J
18.5
c. Given, mass = m, coeff = mu, angle = theta, spring = k, compressed distance =d
let the distance travelled be x
work done by friction = mu*mgcos(theta)*x
initial stored pe in spring = 0.5kx^2
final pe of the system = mgh = mgxsin(theta)
mgxsin(theta) + mu*mgcos(theta)*x = 0.5kx^2
x = sqroot[[mgsin(theta) + mu*mgcos(theta)]/0.5k]
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