(996) Problem 6: A box slides down a plank of length d that makes an angle of wi
ID: 1774418 • Letter: #
Question
(996) Problem 6: A box slides down a plank of length d that makes an angle of with the horizontal as shown. Pk is the kinetic coefficient of friction and , is the static coefficient of friction. ©theexpertta.com 3396 Part (a) Enter an expression for the minimum angle (in degrees) the box will begin to slide. 33% 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. 33% Part (c) For a plank of any length at what angle (in degrees) will the final speed of the box at the bottom of the plank be 85 times the final speed of the box when there is no friction present? Assume ,-0.26.Explanation / Answer
here,
a)
let the minimum angle be theta
for the block to move down the plane
frictional force = gravitational force
us * m * g * cos(theta) = m * g * sin(theta)
solving for theta
theta = arctan(us)
b)
the work done by kinetic frictional force , Wk = uk * m * g * cos(theta) * d
c)
let the angle be theta
final speed when there is no friction , v = sqrt(2 * g * d * sin(theta)) ....(1)
when friction is present
uk * m * g * cos(theta) * d = m * g * sin(theta) * d - 0.5 * m* v'^2 ...(2)
and v' = 0.85 * v ...(3)
from (2) and (3)
uk * m * g * cos(theta) * d = m * g * sin(theta) * d - 0.5 * m* (0.85 * v)^2 ...(4)
from (1) and (4)
uk * m * g * cos(theta) * d = m * g * sin(theta) * d - 0.5 * m* (0.85 * (sqrt(2 * g * d * sin(theta))))^2
0.26 * 9.81 * cos(theta) = 9.81 * sin(theta) - 0.5 * (0.85 * (sqrt(2 * 9.81 * sin(theta))))^2
solving for theta
theta = 43.2 degree
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