Question

Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is μk=0.10.

Answer 1

vₐ = 0.771 ft/s

vb = -1.54 ft/s

Explanation:

Block A:

F = ma

Nₐ = - 60 cos60° = 0

Nₐ = 30 lb

Fₐ = 0.1 X 30 = 3lb

Block B:

F = ma

Nb = - 40 cos30° = 0

Nb = 34.64lb

Fb = 0.1 X 34.64 = 3.464lb

T1 + ∑U = T2

(0 + 0) + 60 sin 60° |Δsₐ| - 40 sin 30° |Δsb| - 3|Δsₐ| - 3.464 |Δsb|

= 1/2 (60/32.2) vₐ² + 1/2 (40/32.2) vb²

and

2vₐ = - vb

On solving, we get

vₐ = 0.771 ft/s

vb = -1.54 ft/s