A body with mass m=1kg and initial velocity Vo=0m/s, slides down an inclined pla

Question

A body with mass m=1kg and initial velocity Vo=0m/s, slides down an inclined plane from point A at a height H1=40m. After passing a horizontal region BC the body elevates on the next inclined plane up to a point D with height H3=? and, then, moves back reaching maximum elevation H2=20m on the first inclined plane. The motion is not affected by friction except for the part between point B and C with length L=20m where there is friction between the body and the surface. Find the coefficient of u in BC and find height H3.

Explanation / Answer

PE = mgh = mgH.

After the block passes through BC, it has energy mgH - u(k)mgL.

At the top, it has energy H, so:

mgH - u(k)mgL = mgH. . . . . . . .(i)

when the block goes back up the first incline:
mgH - 2u(k)mgL = mgH. . . . . . .(ii)

multiplying (i) by 2, we get
2mgH - 2u(k)mgL = 2mgH. . . . . (iii)

[2mgH - 2u(k)mgL] - [mgH - 2u(k)mgL] = 2mgH - mgH
H = 2H - H
H = (H + H)/2
(40 m + 20 m)/2
= 30 m.

From (i):
mgH - 2u(k)mgL = mgH
u(k) = (H - H)/(2L)
(40 m - 20 m)/[2(20 m)]
= 0.50.

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