A copper block slides along a level, frictionless surface with a 70 m/s. It then

Question

A copper block slides along a level, frictionless surface with a 70 m/s. It then encounters a degree incline, which is also frictionless. The surface of the incline is 50.0 m in length. At the base of the incline, the block reaches a second level surface, bu this time with = 0.65. The block eventually comes to rest on the rough surface. Determine the velocity of the block at the instantit has travelled 30 m along the incline. ine the change in temperature of the copper block upon coming to rest given 17 erated is retained by the block.

Explanation / Answer

Wf = -f*s = u*m*g*s

from energy conservation

Ei + Wf = Ef

PEi + KEi + Wf = KEf

m*g*L*sin30 + 0.5*m*vo^2 - u*m*g*s = 0.5*m*v^2

g*L*sin30 + 0.5*vo^2 - u*g*s = 0.5*v^2

v^2 = 2*g*L*sin30 + vo^2 - 2*u*g*s

v^2 = (2*9.8*50*0.5)+(70*70)-(2*0.65*9.8*30) = 5007.8

v = sqrt(5007.8) = 70.76 m/s

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