A box of books weighing 305 N is shoved across the floor of an apartment by a force of 599 N exerted downward at an angle of 27.1° below the horizontal. The acceleration of gravity is 9.8 N. If the coefficient of kinetic friction between box and floor is 0.459, how long does it take to move the box 7.48 m, starting from rest? Answer in units of s.

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aristocles aristocles · Answer 1 · 2019-10-25T20:15:38+02:00

Answer:

[tex]t = 1.32 s[/tex]

Explanation:

As we know that the force is applied at an angle of 27.1 degree below horizontal

So here we have two components of applied force is given as

[tex]F_x = 599 cos27.1[/tex]

[tex]F_x = 533.2 N[/tex]

[tex]F_y = 599 sin27.1[/tex]

[tex]F_y = 272.87 N[/tex]

now we have

normal force due to ground on the box is given as

[tex]F_n = F_y + mg[/tex]

[tex]F_n = 272.87 + 305[/tex]

[tex]F_n = 577.87 N[/tex]

now the friction force on the box is given as

[tex]F_f = \mu F_n[/tex]

[tex]F_f = (0.459)(577.87)[/tex]

[tex]F_f = 265.2 N[/tex]

now the net force in horizontal direction is given as

[tex]F_{net} = F_x - F_f[/tex]

[tex]F_{net} = 533.2 - 265.2[/tex]

[tex]F_{net} = 268 N[/tex]

now the acceleration of the box is given as

[tex]a = \frac{F_{net}}{m}[/tex]

[tex]a = \frac{268}{(305/9.81)}[/tex]

[tex]a = 8.62 m/s^2[/tex]

now the time taken by the box to move the distance of d = 7.48 m

[tex]d = \frac{1}{2}at^2[/tex]

[tex]7.48 = \frac{1}{2}(8.62)t^2[/tex]

[tex]t = 1.32 s[/tex]