A block of mass 2.0 kg enters from the left with an initial speed of 6.0 m/s. Th

Question

A block of mass 2.0 kg enters from the left with an initial speed of 6.0 m/s. The surface is frictionless, with a dip of depth y=0.60 m followed by a rise of height h=0.40 m and horizontal stretch of length d and with a coefficient of friction Uk = 0.25.

b. At the lowest point of the dip, specify the kinetic energy of the block, the change in potential energy of the block, and the work done by gravity on the block.

c. Just before the block enters the frictional region, specify the kinetic energy of the block, the change in potential energy of the block (from its initial value), and the work done by gravity on the block (from the lowest point of the dip).

Explanation / Answer

KE1 = 0.5*m*v1^2 = 0.5*2*6^2 = 36 J

potential energy = U1 = 0

(a)

KE1 = 0.5*m*v1^2 = 0.5*2*6^2 = 36 J

(b)

at the lowest pooint

Kinetic energy = K2

potential energy u2 = -m*g*y

K1 + u1 = k2+u2

36 + 0 = -(2*9.8*0.6)+u2

u2 = 47.76 J <<------answer

change in potential energy = m*g*y = 2*9.8*0.6 = 11.76 J

work done = m*g*h = 11.76 J

(c)

K3 = K1+m*g*h = 36+(2*9.8*0.4) = 43.84 J

dU = m*g*(h) = 2*9.8*(0.4) = 7.84 J

Wg = -m*g*(h+y) = -19.6 J

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