A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.4 s.

Question

a.What is the magnitude of the average net force on the box?

b.You put a 3 kg block in the box, so the total mass is now 9 kg, and you launch this heavier box with an initial speed of 5 m/s. How long does it take to stop?

Answer 1

a = change in v/ change in t = -5/1.4 = -3.57 m/s^2
so
F = m a = 6(-3.57) = - 21.4 Newtons

The friction force will be larger because proportional to the normal force (eight)
F = -21.4 (9/6) = -32.1 Newtons friction force
F = m a
a = -32.1 / 9 = 3.57 m/s^2
so same time, 1.4 seconds

(the mass and the retarding force go u in the same proportion)

Answer 2

Düsseldorf