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Jacob87411
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A m1 = 48.0 kg block and a m2 = 104.0 kg block are connected by a string as in Figure P8.36. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 48.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 48.0 kg block as it moves from A to B, a distance of 20.0 m.
I attached the figure.
Ok so work = change in energy.
W=FD
The forces acting on the block in the X direction are Ft (force tension in the rope) and Ffr (Frictional Force). Ft = 9.8*104kg (Maybe this is wrong?) Force Friction = cos37 * 9.8 * 48 * .25
Fx=Ft-Fr= (9.8*104)-(cos37*9.8*48*.25) = 925N
W=FD=925 * 20 = 18504. This is the change in energy, however the box changes height so there is a change in PE.
Change in PE = mgh = 48*9.8*(sin37* 20) = 5661.
So change in KE = Change in total energy - change in PE=18504 - 5661 = 12843 J.
Thanks for taking time ot help
I attached the figure.
Ok so work = change in energy.
W=FD
The forces acting on the block in the X direction are Ft (force tension in the rope) and Ffr (Frictional Force). Ft = 9.8*104kg (Maybe this is wrong?) Force Friction = cos37 * 9.8 * 48 * .25
Fx=Ft-Fr= (9.8*104)-(cos37*9.8*48*.25) = 925N
W=FD=925 * 20 = 18504. This is the change in energy, however the box changes height so there is a change in PE.
Change in PE = mgh = 48*9.8*(sin37* 20) = 5661.
So change in KE = Change in total energy - change in PE=18504 - 5661 = 12843 J.
Thanks for taking time ot help