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/χgμequ 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

Explanation:

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

f=ma newton

f=100x2.5=250n

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