Phy 201
Your 'cq_1_11.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Answer the following based on Newton's Second Law:
• How much net force is required to accelerate a 12 kg mass at 3 m/s^2?
answer/question/discussion: ->->->->->->->->->->->-> : We are given the mass and the acceleration of the object. Knowing the equation Fnet = m * a, we can plug these quantities in to get Fnet = 12kg * 3m/s^2 = 36 Newtons.
• What would be the acceleration of a 4 kg mass subject to a net force of 20 Newtons?
answer/question/discussion: ->->->->->->->->->->->-> : Rearranging this above equation around, we get a = Fnet/m. We are given the Fnet as 20N and the mass as 4kg. So, a = 20N / 4kg = 5m/s^2.
• If you exert a force of 20 Newtons on a 10-kg object and it accelerates in the direction of your force at 1.5 m/s^2, then how do you know there are other forces acting on the object besides your own?
answer/question/discussion: ->->->->->->->->->->->-> : Because using the given mass and acceleration, the total Fnet should be Fnet = 10kg * 1.5m/s^2 = 15 Newtons. So if the Net Force is 15 Newtons, and the force exerted on the object is 20 Newtons, there must be 5 Newtons of force working in the opposite direction of your applied force.
• What is the total of all those forces and in what direction does this total act?
answer/question/discussion: ->->->->->->->->->->->-> : I can only imagine that you are referring to the above problem. In that case….. The total net force (as I said above) was 15 Newtons, in the direction of the exerted force. You are pushing a 10kg object at 1.5m/s^2, but you are exerting a force of 20 Newtons on this object. This can only lead to the conclusion that there must 5 Newtons of force working against you. In that case, the Fnet would = 15 Newtons, which correlates to the 10kg object being pushed as 1.5m/s^2 (Fnet = 10kg * 1.5m/s^2 = 15 N).
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About 20 minutes...
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&#Your work looks very good. Let me know if you have any questions. &#