tacotime
  • tacotime
A 4.30 kg box is pushed 1.20 m with a 36 N force. What is the final speed of the box if the coefficient of kinetic friction is 0.30?
Physics
chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
on your box are acting two forces, namely: the driving force F=20 N and the friction force: R=\mu*m*g=0.3*4.30*9.81=...N Now the work done by those forces has to be equal to the kinetic energy change, so we can write this equation: \[Fd - Rd = \frac{1}{2}m{v^2}\] where \(v\) is the requested final speed, and \(d\) is the distance traveled by your box, namely \(d=1.2 \: m\)
Michele_Laino
  • Michele_Laino
of course, \(m\) is the mass of your box
tacotime
  • tacotime
thank you so much

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