A box slides across a frictionless floor with an initial speed v = 1.6 m/s. It encounters a rough region where the coefficient of friction is µk = 0.5.
x=0.26m
*If instead the strip is only 0.0784 m long, with what speed does the box leave the strip?
v f =?????????????
Physics 101: box/friction, someone please help. This is the one problem I can't complete. I figured out our first x, what is Vf?
2014-10-08 6:16 pm
回答 (1)
2014-10-08 6:30 pm
By the conservation of energy, we know that total energy TE = PE + KE + QE = constant; where PE is stored energy, KE kinetic, and QE is other like heat or sound.
So your box starts with TE = KE = 1/2 mU^2 and ends up with TE = ke + qe after passing over the strip. qe = kNS is the work done by the friction to sap some of the KE and slow down the box. k = .5 and N = mg is the normal force based on the weight W = mg of the box.
So from the COE law we have TE = 1/2 mU^2 = 1/2 mv^2 + kmgS = TE We solve for v = sqrt(U^2 - k2gS) = sqrt(1.6^2 - .5*2*9.8*.26) = .11 m/s ANS.
If the slide distance is S = .0784, put that value into v = sqrt(U^2 - k2gS) and solve. You can do the math. ANS.
So your box starts with TE = KE = 1/2 mU^2 and ends up with TE = ke + qe after passing over the strip. qe = kNS is the work done by the friction to sap some of the KE and slow down the box. k = .5 and N = mg is the normal force based on the weight W = mg of the box.
So from the COE law we have TE = 1/2 mU^2 = 1/2 mv^2 + kmgS = TE We solve for v = sqrt(U^2 - k2gS) = sqrt(1.6^2 - .5*2*9.8*.26) = .11 m/s ANS.
If the slide distance is S = .0784, put that value into v = sqrt(U^2 - k2gS) and solve. You can do the math. ANS.
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