Question

(7%) Problem 13: A box slides down a plank of length d that makes an angle of θ with the horizontal as shown. μk is the kinetic coefficient of friction and us is the static coefficient of friction Otheexpertta.comm 33% Part (a) Enter an expression for the minimum angle (in degrees) the box will begin to slide. θmin-atan(με) Correct! 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 θ is large enough that it will move down the plank. wnc-μk m g cos(θ) d X Attempts Remain Feedback: is available 33% Part (c) For a plank of any length, at what angle θ (in degrees) will the final speed of the box at the bottom of the plank be 0.75 times the final speed of the box when there is no friction present? Assume μ,-0.36. Grade Summary Deductions .% Potential 100%

Answer 1

B)

d = distance traveled

Perpendicular to incline, force equation is given as

F_n = mg Cos$\theta$

kinetic frictional force is given as

f_k = u_k F_n

f_k = u_kmg Cos$\theta$

Work done by kinetic frictional force is given as

W_nc = f_k d Cos180

W_nc = - u_kmgd Cos$\theta$

c)

when there is friction :

h_i = initial height from where the block starts = d Sin$\theta$

v = final speed at the bottom

using conservation of energy

mg h_i - W_nc = (0.5) m v²

mg (d Sin$\theta$) - u_kmgd Cos$\theta$ = (0.5) m v²

2g (d Sin$\theta$) - 2u_kgd Cos$\theta$ = v²

v = sqrt(2g (d Sin$\theta$) - 2u_kgd Cos$\theta$)                                             eq-1

When there is no friction :

v' = final speed at the bottom

using conservation of energy

mg h_i = (0.5) m v'²

v' = sqrt(2g d Sin$\theta$)                                                     eq-2

it is given that

v = 0.75 v'

Using eq-1 and eq-2

sqrt(2g (d Sin$\theta$) - 2u_kgd Cos$\theta$) = (0.75) sqrt(2g d Sin$\theta$)

sqrt(Sin$\theta$ - u_k Cos$\theta$) = (0.75) sqrt( Sin$\theta$)

sqrt(Sin$\theta$ - (0.36) Cos$\theta$) = (0.75) sqrt( Sin$\theta$)

Sin$\theta$ - (0.36) Cos$\theta$ = (0.5625) Sin$\theta$

(0.4375) Sin$\theta$ = (0.36) Cos$\theta$

tan$\theta$ = 0.823

$\theta$ = 39.5 deg