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 Oct 29th 2018, 01:40 PM #1 Junior Member   Join Date: Oct 2018 Posts: 4 Newtons laws of motion A rope pulls a box of mass 2 kg along a horizontal floor with a constant speed. The friction force opposing the motion is 12 N to the left. Determine the magnitude and direction of the force pulling the box. Is it 12 N to the right because of N's 3rd law?
Oct 29th 2018, 02:02 PM   #2
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 Originally Posted by coolbat14 A rope pulls a box of mass 2 kg along a horizontal floor with a constant speed. The friction force opposing the motion is 12 N to the left. Determine the magnitude and direction of the force pulling the box. Is it 12 N to the right because of N's 3rd law?
Not by Newton's 3rd.

Sketch a free body diagram. There is the applied force (F) on the box going to the right (friction (f) is always against the direction of motion.) Since the box is moving at constant speed the acceleration is zero and hence the net force on the box is 0 N. Net force is $\displaystyle \sum F = F - f = 0$ and thus F = -f.

-Dan
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 Oct 29th 2018, 02:11 PM #3 Junior Member   Join Date: Oct 2018 Posts: 4 Thanks this helps a lot!
 Oct 29th 2018, 02:22 PM #4 Junior Member   Join Date: Oct 2018 Posts: 4 I have another question.. A mass of 4kg moving with a velocity of 6m/s suddenly experiences a force and accelerates for 7s until its velocity is 9m/s in the same direction. Calculate the resultant force on the mass. Do I use F = (mv-mu)/t to get = (4*9 - 4*6)/7 = 1.7N in the same direction?
 Oct 30th 2018, 03:19 PM #5 Senior Member   Join Date: Aug 2010 Posts: 403 Another way to calculate that is to say that the object has accelerated from speed 6 m/s to 9 m/s so has a change of speed 9- 6= 3 m/s. It did that in 7 s so that is an acceleration of 3/7 m/s^2. Since "force= mass times acceleration" the Force is (4 kg)(3/7 m/s^2)= 12/7 Newtons. (Rounded to one decimal place that is 1.7 but 12/7 is exact.) I am just using m(u- v)/r rather than your (mu- mv)/t.

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