here,
mass of book ,m1 = 1.5 kg
mass of cup , m2 = 0.45 kg
uk = 0.2
the accelration , a = net force/effective mass
a = ( m1 * g * sin(theta) + m2 * g + uk * m1 * g * cos(theta))/( m1 + m2)
a = ( 1.5 * 9.81 * sin(20) + 0.45 * 9.81 + 0.2 * 1.5 * 9.81 * cos(20))/( 1.5 + 0.45)
a = 6.26 m/s^2
let the distance travelled before stopping be s
v^2 - u^2 = 2 * a * s
0 - 3.3^2 = - 2 * 6.26 * s
s = 0.87 m
b)
the downward force , Fd = m1 * g * sin(theta) + m2 * g
Fd = 1.5 * 9.81 * sin(20) + 0.45 * 9.81 N
Fd = 9.44 N
the static frictional force , ffs = us * m1 * g * cos(theta)
ffs = 1.5 * 0.5 * 9.81 * cos(20) = 6.91 N
as Fd > ffs
the block will slide downwards
premouS11 Problem 7.41 Part A The 1.5 kg physics book in figure is connected by a...
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