A 3.0kg object is sliding on the table with a velocity of 2.6 m/s to the right.

Question

A 3.0kg object is sliding on the table with a velocity of 2.6 m/s to the right. a 6.0N friction force acts on the object. That force is directed to the right. In order to keep the object moving with a constant velocity, we must exert a 6.0N force directed to the right. As soon as we stop exerting this force, the object will slow down with an acceleration of magnitude ( either a) 0, b) 2.0, c) 6.0) to the left. The object will come to rest in how many seconds and an additional distance of how much to the right?

Explanation / Answer

The frictional force will act to the left, and in absence of external 6N force to the right, the object will slow down and stop.

a= acceleration= Force/ mass = 6N/ 3kg= 2 m/s^2 to the left

so, b) acceleration= 2m/s^2 to the left

Using the relation v= u+at {where v= final speed= 0; u= initial speed= 2.6m/s; a=-2m/s^2 }

So, 0= 2.6-2t

or, t= 2.6/2= 1.3

So, object will come to rest in 1.3 seconds.

Now, using the relation s= ut+0.5at^2 { where u= 2.6; a=-2; t=1.3; s= additional distance covered to the right }

So, s= 2.6*1.3+0.5*(-2)*1.3^2

Or, s= 3.38-1.69= 1.69m

So, additional distance of 1.69m is covered to the right

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