Find speed using the work-energy theorem

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Homework Statement


A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.25. What is the final speed of the crate?

Homework Equations


Work energy theorem ##\Delta W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ##

The Attempt at a Solution


- First I'll calculate the speed at the end of the first fifteen meters (when there is no acceleration).
##F \cdot d = (350)(15) = 5250 = ## work
##= \Delta E_k = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ##
Because the initial velocity is zero, we see that ## = \frac{1}{2}mv_f^2 ##
##10500 = mv^2 ## so ##v_f = 10.458 ## m/s.

- Now I'll calculate the answer (speed at 30 meters of displacement).
Due to the presence of friction, kinetic and potential energies are not the only energies present.
##E_K + E_P + E_f = ## where the subscript f refers to friction
##\frac{1}{2}mv_0^2+mgh+F_N\mu kl = \frac{1}{2}mv_f^2+mgh+F_N\mu kl ##
##v_0 = 10.458##
##5250+0+0 = \frac {1}{2}mv_f^2 + 0 + 3528 ##
##v_f = 5.98## m/s
The correct answer is twice what I got. What did I do wrong?
 

Answers and Replies

  • #2
robphy
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(There is acceleration during the first 15 m.... there is no friction, however.)
During the next 15m, is it still being pulled by that same applied force of 350N?
 
  • #3
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Homework Statement


A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.25. What is the final speed of the crate?

Homework Equations


Work energy theorem ##\Delta W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ##

The Attempt at a Solution


- First I'll calculate the speed at the end of the first fifteen meters (when there is no acceleration).
##F \cdot d = (350)(15) = 5250 = ## work
##= \Delta E_k = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ##
Because the initial velocity is zero, we see that ## = \frac{1}{2}mv_f^2 ##
##10500 = mv^2 ## so ##v_f = 10.458 ## m/s.

- Now I'll calculate the answer (speed at 30 meters of displacement).
Due to the presence of friction, kinetic and potential energies are not the only energies present.
##E_K + E_P + E_f = ## where the subscript f refers to friction
##\frac{1}{2}mv_0^2+mgh+F_N\mu kl = \frac{1}{2}mv_f^2+mgh+F_N\mu kl ##
##v_0 = 10.458##
##5250+0+0 = \frac {1}{2}mv_f^2 + 0 + 3528 ##
##v_f = 5.98## m/s
The correct answer is twice what I got. What did I do wrong?

##W = W_1 + W_2 ##

W_1 (first 15 meters) and W_2 (second 15 meters)

## W=\Delta E_K=\frac{1}{2}m v_f^2## (given that v_0 = 0 )

So:

##\frac{1}{2}mv_f^2=W_1 + W_2 = F.d + (F-R).d##

Being ##m=96, F=350, d=15, R=0.25*m*9.8##


So you have:

##v_f = \sqrt{\frac{2(2F-R)d}{m}}##
 

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