A box of books weighing 320 N is shoved across the floor of an apartment by a force of 430 N exerted...?

Question

A box of books weighing 320 N is shoved across the floor of an apartment by a force of 430 N exerted downward at an angle of 35.0° below the horizontal. If the coefficient of kinetic friction between box and floor is 0.57, how long does it take to move the box 4.10 m, starting from rest?

Answer 1

The force of the shove has two components:
Horizontal: F cos theta
Down: F sin theta

The weight is just w down

So the normal force is:
w + F sin theta (up)

So the friction force opposing the motion is
mu (w + F sin theta)

So the net force is:
push - friction
= F cos theta - mu (w + F sin theta)

The mass is just w/g

So acceleration is given by force over mass:
a = Fg/w cos theta - mu (g + Fg/w sin theta)

The distance traveled from rest is given by:
d = 1/2 at^2
so t = sqrt (2d/a)
= sqrt (2d / (Fg/w cos theta - mu (g + Fg/w sin theta)))

They give you the force, the weight, the angle, the friction coefficient, and the distance. You know g. Plugnchug.

Answer 2

Ff = (0.57)(320 + 430sin35°)
Fp = 430cos35°
a = (430cos35° - (0.57)(320 + 430sin35°))(9.80665)/320
t = √((2)(4.1)(320)/((430cos35° - (0.57)(320 + 430sin35°))(9.80665)))
t = √(2,624/((3,454.24928541 - 1,788.73296 - 1,378.654)))
t ≈ √(2,624/286.86232541)
t ≈ 3.0244 sec.