Alessandraidn
Answered

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a force of 300N[S] is applied to a 50kg box on a rough floor. The frictional force is 200N. The acceleration is 2m/s^2.

if the 300N force is removed after 4.0s, how long will it take the box to come to rest?

please show ALL your work!

Sagot :

We want to know how far this box will move after the force stops. This is called the displacement and it can be calculated using:

[tex]Displacement = Final\ Velocity \times Time[/tex]

We do not know the final velocity of the box however this can be calculated using:

[tex]Final\ Velocity = Initital\ Velocity + (Acceleration \times Time)[/tex]

When we put our values into this equation we get:

[tex]Final\ Velocity = 0 + (2m/s^2 \times 4.0s)[/tex]
[tex]Final\ Velocity = 8m/s[/tex]

We can put this value into the Displacement equation to get

[tex]Displacement = 8m/s \times 4s[/tex]
[tex]Displacement = 32m[/tex]

This means that the total difference in space between where the box stopped receiving a force and came to halt is 32m. Presuming the floor was completely flat, then the only way the box could move is forwards. This means that there is a 32m movement forwards between the two places which means the box took 32m to come to rest.
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