Fnet = ma Ff = -0.3375/0.08 = -4.219N

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The discussion revolves around calculating the force of friction acting on a 0.08 kg box shot across the ground using a slingshot with a spring constant of 60 N/m. The initial elastic potential energy (EPE) of the slingshot is calculated as 0.3375 J. The work done by the frictional force is equal to the negative of this energy, leading to a calculation of friction force using the formula W = F * D. Participants question the accuracy of the calculations, particularly regarding the halving of values. The conversation emphasizes the need to verify the initial energy calculation to ensure accurate results.
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Homework Statement



You shoot a 0.08 kg box across the ground with 60 N/m slingshot, stretched back 0.15m. The box slides 4.22m before coming to a stop. What is the force of friction acting on the box as it slides across the ground?

Homework Equations



EPE = 1/2kx^2

The Attempt at a Solution



E1 = 1/2(60)(0.15)^2
E1 = 0.3375J
W = 0 - 0.3375J = -0.3375J
 
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CollegeJunior said:

The Attempt at a Solution



E1 = 1/2(60)(0.15)^2
E1 = 0.3375J
W = 0 - 0.3375J = -0.3375J

Right so all the work done by the slingshot will be opposided by the work done by the frictional force.

How would you calculate the work done by the friction force F with a distance of 4.22 m ?
 
rock.freak667 said:
Right so all the work done by the slingshot will be opposided by the work done by the frictional force.

How would you calculate the work done by the friction force F with a distance of 4.22 m ?

W = F * D, so F = W / D? I got around -0.08 N for force. Is that the Force of Friction?
 
CollegeJunior said:
E1 = 1/2(60)(0.15)^2
E1 = 0.3375J
You seem to have halved twice over.
 
haruspex said:
You seem to have halved twice over.

?
 
CollegeJunior said:
?

Verify your calculation of E1.
 
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