You put a 3 kg block in the box, so the total mass is now 8 kg, and you launch this heavier box with an initial speed of 4 m/s. How long does it take to stop

0.7 secs

Explanation:

In this question, the speed does not change as the mass changes. So we can use

Δt=Δ∨x/χgμ ... equ 1

To stop, the final speed will be 0

Therefore,

Δvx=vf-vt

Δvx=0-4m/s

= - 4m/s

Now substitute the various values in equ 1

Δt=Δ∨x/χgμ

Δt = - 4m/s / (9.8m/s∧2) (0.6)

Δt=0.7 secs

The box stops at zero speed.

Final Velocity = 0, Initial speed (s) = - 4 m/s

Therefore = change in velocity = Vf - Vi. (0 m/s - 4 m/s) = - 4 m/s

Change in velocity = - 0.4 m/s

Gravity g = 9.8 m/s^2

Mass = 0.8 g

-4 ms divided by 9.8 ms^2 * (0.8) = 0.51 s

It takes 0.51 seconds to stop