anonymous
  • anonymous
A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5N acts on it for 4 sec. What will be the distance of the block from the point where the force started acting. A. 10m. B. 8m C.6m. D.2m
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@irishboy123
anonymous
  • anonymous
@arindameducationusc
anonymous
  • anonymous
@michele_laino

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Michele_Laino
  • Michele_Laino
the situation of your problem can be summarized as follows: |dw:1446396527186:dw| namely the motion of the block is a motion which is contained inside the horizontal plane \(x,y\). Gravity is acting perpendicularly with respect to that plane, therefore, it is represented by a vector which is not contained inside such motion plane. Now on the block is acting the external force \(F\), so the acceleration \(a\), of such block is: \(a=F/m\), finally the space \(S\) traveled by the block, under the action of force \(F\), namely inside the interval time \((0,4)\), is: \[\Large S = \frac{1}{2}\frac{F}{m}{\tau ^2} = ...?\] where \(\tau=4\) seconds
Michele_Laino
  • Michele_Laino
and, of course, \(m=5\) Kg
IrishBoy123
  • IrishBoy123
and then don't forget to add in the constant motion in other direction, ie |dw:1446404590846:dw| so Pythagoreas it....
anonymous
  • anonymous
I think we should not consider the force which is acting perpendicular to the plane becaise cos 90=0 is'nt it???
IrishBoy123
  • IrishBoy123
well, your question is ambiguous. michele's drawing presumes that it acts along the plane in the y direction. if it acts in the z directions, into the plane, the question becomes somewhat pointless.
anonymous
  • anonymous
Read the question carefully....it is just a simple 1 dimensional mechanics problem....a box is moving with a velocity v and a perpendicular force acts on it for t secs and we just have to find S the distance .....as simple as that

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