Given. Block A: weight 200 lb, on a frictionless 30° incline. Block B: weight 350 lb, hangs from a movable pulley. Cable path: A → fixed pulley at the top of the incline → movable pulley (carrying B) → fixed ceiling anchor. Pulleys massless and frictionless.
Given data
Quantity
Value
Weight of block A
200 lb
Weight of block B
350 lb
Incline angle
30°
Distance A travels
5 ft
Find. a) The velocity of block A after moving 5 ft along the incline; b) the tension in the (single, continuous) cable.
Figure 5 — block A on the incline, connected over a fixed pulley to a movable pulley carrying block B, which is also anchored to the ceiling.
Approach. Because the movable pulley carrying B has two supporting cable segments (one to the fixed top pulley, one to the fixed ceiling anchor), each inch B descends lengthens BOTH of those segments by one inch, so the single segment running to A must shorten by twice as much: $v_A = 2v_B$. Apply the work-energy theorem to the whole system, then use the resulting (constant) acceleration in Newton's second law for block A to get the cable tension.
Kinematic constraint. With B descending as A moves up the incline: $s_A = 2 s_B \Rightarrow v_A = 2v_B,\ a_A=2a_B$. For $s_A=5$ ft, $s_B = 2.5$ ft.
Work-energy theorem, system. Gravity does negative work on A (rising along the incline) and positive work on B (descending):
$$U_{1\to2} = -W_A\sin30^\circ\,s_A + W_B\,s_B = -200(0.5)(5)+350(2.5) = 375\ \text{ft-lb}$$
Kinetic energy and solve for velocity. With $v_A=2v_B$ and $m=W/g$ ($g=32.2\ \text{ft/s}^2$):
$$U_{1\to2} = \tfrac12 m_A v_A^2+\tfrac12 m_B v_B^2 = \tfrac12 v_B^2\left(4m_A+m_B\right)$$
$$v_B = \sqrt{\dfrac{375}{\tfrac12(4m_A+m_B)}} = 4.583\ \text{ft/s} \quad\Rightarrow\quad v_A = \boxed{9.17\ \text{ft/s}}$$
Tension — constant acceleration. Since all forces are constant, $v_A^2=2a_As_A$:
$$a_A = \dfrac{v_A^2}{2s_A} = \dfrac{9.165^2}{2(5)} = 8.40\ \text{ft/s}^2$$
Newton's second law along the incline for A (tension T pulls up-slope):
$$T - W_A\sin30^\circ = m_A a_A \Rightarrow T = \dfrac{200}{32.2}(8.40)+200(0.5) = \boxed{152.2\ \text{lb}}$$
Check on block B. Two cable segments support B, so the net upward force is $2T$; with $a_B=a_A/2=4.20\ \text{ft/s}^2$ downward:
$$W_B - 2T = m_B a_B \Rightarrow 350-2(152.2) = 45.6\ \overset{?}{=}\ \dfrac{350}{32.2}(4.20)=45.7$$
Agrees to rounding (residual < 10⁻&sup9; lb in the exact Python solve) — self-consistent.
