Abox of books weighing 305 n is shoved

across the floor by a force of 516 n exerted

downward at an angle of 32° below the hori-

zontal.

the acceleration of gravity is 9.81 m/s.

if her between the box and the floor is 0.59,

how long does it take to move the box 5.82 m,

starting from rest?

3.1 s

Explanation:

from the question we are given the the following:

Weight of books (Fb) = 305 N

Push force (Fp) = 516 N

distance (s) = 5.82 m

Angle of force exerted = 32 degrees

acceleration due to gravity (g) = 9.81 m/s^{2}

coefficient of friction Uk= 0.59

time (t) = ?

mass of the book (m) = weight / (g) =  305 / 9.8 = 31 kg

lets first get the net force from the summation of vertical forces

Fnet = Fpush + Fgravity

F net = 516 + 305sin32

F net = 677.6 N

now lets get the acceleration from the summation of the horizontal forces

Fpcos32 - friction force = m x a

Fpcos32 - (Uk x F net) = m x a

516cos32 - (0.59 x 677.6) = 31 x a

37.8 = 31a

a =  1.22m/s^{2}

now that we have our acceleration we can get the time from the equation of motion

s = ut + o.5at^{2}

u ( initial velocity ) = 0 because the box was initially at rest

5.82 = (0 x t ) + (0.5 x 1.22 x t^{2})

5.82 = 0.61t^{2}

t = 3.1 s

the inner core.

good luck: )

klicker