An object with a mass of 7 kg is pushed along a linear path with a kinetic friction coefficient of mu_k(x)= 4+secx. How much work would it take to move the object over x in [(-5pi)/12, (pi)/4], where x is in meters?
1 Answer
w_"push" = ul(7((8pi)/3vecg + (2pi)/3vecgsecx)) "J"
x in [-(5pi)/12, pi/4]
vecg > 0
And the work itself is done from the perspective of the worker.
Well, I would use a conservation of energy approach, assuming the object has no kinetic energy (i.e. it does not keep moving if we stop pushing). We just need to overcome the barrier of kinetic friction over a distance
DeltaE = 0 = w_(vecF_k) - w_"push" where:
w_"push" is the work one would have to do to push the objectpi/4 + (5pi)/12 = (2pi)/3 meters forward.w_(vecF_k) = vecF_kDeltax is the counteracting work done by the kinetic friction.vecF_k = mu_kvecF_N is the friction force, andvecF_N is the normal force.mu_k is the coefficient of kinetic friction.
Here, we do the work from the perspective of ourselves, so
As for
Including the horizontal sum of the forces:
sum_i vecF_(x,i) = vecF_"push" - vecF_k = mveca_x
We realize that by not knowing the acceleration, we couldn't solve for
We do, however, find a use in including the vertical sum of the forces:
sum_i vecF_(y,i) = vecF_N - vecF_g = vecF_N - mvecg = 0 ,where our convention is that
vecg > 0 (and the negative is in the subtraction sign), we have:
We now have an expression for
color(blue)(w_"push") = vecF_kDeltax
= mu_kvecF_N Deltax
= (4 + secx)(mvecg)((2pi)/3)
= (4 + secx)("7 kg" xx g" m/s"^2)((2pi)/3 "m")
= color(blue)(7((8pi)/3g + (2pi)/3gsecx)) color(blue)("J")
The work is seen to be a function of the position (but only in
The initial push should then be hardest, the middle of the distance should be easiest, and then near the end of the distance should be slightly harder.