The 40kg block is moving to the right with a speed of 1.5m/s when it is acted upon by the forces F 1 and F . These forces vary in the manner shown in the graph. Find the velocity of the block at t =

Question

The 40kg block is moving to the right with a speed of 1.5m/s when it is acted upon by the forces F 1 ​ and F . These forces vary in the manner shown in the graph. Find the velocity of the block at t = 6 s . Neglect friction and mass of the pulleys and cords. (Velocity in m/s)

TestHub · Accepted Answer

The right-hand cord system is a movable-pulley cascade, so the block is pulled right with $4 F_{1}$ ; the left cord pulls it left with $F_{2}$

Impulse of $4 F_{1}$ from $t = 0$ to 6 s: $4 \times (10 \times 2 + 20 \times 2 + 30 \times 2) = 480$ N-s

Impulse of $F_{2}$ (acts only from $t = 4$ to 6 s): $40 \times 2 = 80$ N-s

Net impulse $= 480 - 80 = 400$ N-s $= m Δ v \Rightarrow Δ v = \frac{400}{40} = 10$ m/s

$v = 1.5 + 10 = 11.5 \approx 12$ m/s

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