iPad e session.masteringphysics.com Physics 101 Summer 2018 Manal& Chapter 5 Pro

Question

iPad e session.masteringphysics.com Physics 101 Summer 2018 Manal& Chapter 5 Problem 5.75 You place a book of mass 3 0p ikg against a vertical wall You aply a constant force F N and the force is at angle of 60.0 above the horizontal The coefficient of kinetic friction between the book and the wall is 0.300. F to the book, where F-90.0 60 Part A If the book is initially at rest, what is its speed after it has traveled 0.400 m up the wal, Express your answer with the appropriate units. ? ? u= 1 Value Units Submit Request Answer Provide Feedback Next >

Explanation / Answer

SOLUTION :

we have

mass of book 3 k.g

friction coefficient, u = 0.30

angle of force acting on book, theta = 60 degree as shown in question's figure

gravitational force acting downward = m*g = 3*9.8 = 29.4 N

force acting on book, F = 90 N

frictional force acting downward, f = u*N1 where N1 is normal force and N1 is given by

N1 = F*cos(theta) = 90*cos(60) = 45 N ............( forces acting perpendicular to the motion of book)

force acting on book along the motion = F*sin(theta) - m*g -N1 = 90*sin(60) - 29.4 - 45

= 77.9423 - 29.4 - 45 = 3.5423 N

therefore acceleration of book along motion,a = 3.5423 / 3 = 1.18 m/s2

Now from 3rd equation of motion

v2 = u12 + 2*a*s

here u1 = 0, a = 1.18 m/s2, s = 0.40 m , v = velocity at distance of 0.40 m

therefore v2 = 0 + 2*1.18*0.40 = 0.9446

v = 0.972 m/s

hence, velocity of book at a distance of 0.40 m is v = 0.972 m/s ANSWER v = 0.972 m/s

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