(9%) Problem 11 : A box slides down a plank of length d that makes an angle of w
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Question
(9%) Problem 11 : A box slides down a plank of length d that makes an angle of with the horizontal as shown. k is the kinetic coefficient of friction and u, is the static coefficient of friction. Ctheexpertta.com ee 50% Part (a) Enter an expression for the minimum angle (in degrees) the box will begin to slide. 50% Part (b) Enter an expression for the nonconservative work done by kinetic friction as the block slides down the plank. Assume the box starts from rest and is large enough that it will move down the plank. nc Potential9 | (171819 acotanqus) | | | atan(us) cos(0) sin(0) | | cos(a) sin(a) Attempts remain (5% per attempt cos() sin() Feedback I give up! Submit Hint Feedback: deduction per feedback. Hints: 2% deduction per hint. Hints remaining: 2Explanation / Answer
A) tan(theta) = mu_s
theta = tan^(-1)(mu_s)
b) work done by the frictional force is W = change in mechanical energy
W = Final mechanical energy - initial mechanical energy
W = (0.5*m*v^2)-(m*g*h)
work done by the Net force = change in kinetic energy
Work done by the gravitational force + frictional force = 0.5*m*v^2
m*g*sin(theta)*d + Wf = 0.5*m*v^2
Net Force = Fnet = m*a
m*g*sin(theta) - (mu_k*m*g*cos(theta)) = m*a
accelaration is a = g[sin(theta)-mu_k*cos(theta)] = a
using kinematic equations
v^2 - vo^2 = 2*a*S
v^2 = 2*g[sin(theta)-mu_k*cos(theta)]*d
then
m*g*sin(theta)*d + Wf = 0.5*m*v^2
wf = [0.5*m*2*g[sin(theta)-mu_k*cos(theta)]*d] - [m*g*sin(theta)*d]
Wf = -mu_k*m*g*cos(theta)*d
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