Question

A 20-kg box rests on the flat floor of a truck. The coefficients of friction between the box and floor are μs = 0.15 and μk = 0.10. The truck stops at a stop sign and then starts to move, with an acceleration of 2.0 m/s².
a) If the box is 5.0 m from the rear of the truck when it starts, how much time elapses before it falls off the rear of the truck? 3.13 s
b) How far does the truck travel in this time?

Answer 1

Answer:

a) 3.16s

b) 10m

Explanation:

Let g = 10m/s2. Gravity and normal forces acting on the box:

N = G = mg= 20*10 = 200 N

The static friction acting on the box is

$F_s = N\mu_s = 200 * 0.15 = 30N$

Kinetic friction acting on the box is:

$F_k = N\mu_k = 200 * 0.1 = 20N$

When the truck accelerates from rest, this acceleration would generate an inertia force on the box

$F_i = a_im = 2 * 20 = 40N$

Since$F_i > F_s$ the box is definitely moving with respect to the truck with a net force of

$F_b = F_i - F_k = 40 - 20 = 20N$

So then the net acceleration of the box with respect to the truck is

$a_b = F_b / m = 20 / 20 = 1 m/s^2$

a) the distance to the rear truck is 5m, then the time it takes for the box to accelerate and reach that point is

$d = \frac{a_bt^2}{2}$

$t^2 = \frac{2d}{a_b} = \frac{2*5}{1} = 10$

$t = \sqrt{10} = 3.16s$

b) The distance travelled by the truck when it starts from rest:

$d_t = \frac{a_t t^2}{2} = \frac{2 * 3.16^2}{2} = \frac{20}{2} = 10 m$