A force of 300 N [S] is applied to a 50 kg box on a rough floor. The frictional force is 200 N. The acceleration is 2 m/s².
If the 300 N force is removed after 4.0 seconds, how long will it take the box to come to rest?
Please show all your work!
Answer (3)
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:
D i s pl a ce m e n t = F ina l V e l oc i t y × T im e
We do not know the final velocity of the box however this can be calculated using:
F ina l V e l oc i t y = I ni t i t a l V e l oc i t y + ( A cce l er a t i o n × T im e )
When we put our values into this equation we get:
F ina l V e l oc i t y = 0 + ( 2 m / s 2 × 4.0 s ) F ina l V e l oc i t y = 8 m / s
We can put this value into the Displacement equation to get
D i s pl a ce m e n t = 8 m / s × 4 s D i s pl a ce m e n t = 32 m
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.
To find out how long it will take for the box to come to rest after the 300N force is removed, we need to determine the deceleration of the box. We can use Newton's second law, which states that the net force acting on an object is equal to mass times acceleration (F = ma). Since the box is coming to rest, the net force is equal to the frictional force.
So, we have Fnet = Ffriction = 200N and m = 50kg. Rearranging the equation, we can solve for the deceleration (a): a = Fnet/m. Plugging in the values, we get a = 200N/50kg = 4m/s^2.
Now we can use the equation for uniformly decelerated motion, v = u + at, where v is the final velocity (0m/s), u is the initial velocity (2m/s), a is the deceleration (-4m/s^2), and t is the time taken. Rearranging the equation, we get t = (v - u)/a. Plugging in the values, we get t = (0m/s - 2m/s)/-4m/s^2 = 0.5s. Therefore, it will take the box 0.5 seconds to come to rest after the 300N force is removed.
The box will take 2 seconds to come to rest after the force is removed. This is calculated by determining the initial velocity achieved during the force application and then considering the deceleration caused by friction. The calculations show the box decelerates at 4 m/s² after the force is removed, leading to a stop in 2 seconds.
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