Calculating Acceleration and Work for a Box on a Horizontal Surface

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SUMMARY

The discussion focuses on calculating the acceleration and work done on a 125 N box being pulled with a force of 60 N at an angle of 42 degrees, while experiencing a frictional force of 15 N. The net force acting on the box is determined to be 45 N, leading to an acceleration of 0.237 m/s². Additionally, the work done to move the box over a distance of 5 m is calculated using the formula W = F × d, where F is the net force. The calculations emphasize the importance of correctly identifying mass and force in physics equations.

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  • Understanding of Newton's Second Law (F=ma)
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  • Familiarity with work-energy principles
  • Basic trigonometry for resolving forces at angles
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of forces acting on objects, particularly in mechanics and engineering contexts.

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Homework Statement


A 125 N box is pulled along the floor with a force of 60 N acting at an angle of 42 degrees to the horizontal. If the force of friction on the box is 15 N,
A. what is the acceleration of the box?
B. How much work is done to move the box over a distance of 5 m?


Homework Equations





The Attempt at a Solution


TX = Tcos0
TX = 60Ncos42
TX = 44.6N

fnet=60N - 45N
fnet = 45N

fnet= TX - Ff = ma
45N = 44.6N - 15N = (125N)(a)
45N = .237 = a
a=189.9m/s^2
 
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fnet= TX - Ff = ma
45N = 44.6N - 15N = (125N)(a)
45N = .237 = a
a=189.9m/s^2

you should only have 44.6-15N= ma
Where are you getting the 45N on the left hand side of the equation?
Also, remember that the "m" in F=ma is the MASS of the object...What are units of mass? 125N is not a mass.
 

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