Question

In the adjacent figure, $AB$ is a fixed horizontal rod. A sleeve $S$ of mass $m = 9$ kg is free to move along the rod. One end of a light and inextensible string which is passing over a fixed pulley is connected to the sleeve. A block of mass $M = 25$ kg is attached to the other end which keeps the other part vertical. The system is set free from rest with the sleeve in position I. Calculate the speed of sleeve in the nearest integer value when it reaches the position II. Ignore any friction.

Answer 1

4 m/s

Answer 2

We solve the problem by applying energy conservation and the kinematic constraint of the string.

Let

• $m = 9$ kg (sleeve mass),
• $M = 25$ kg (block mass),
• $v$ = speed of sleeve (along the horizontal rod at II),
• $v_M$ = speed of mass M (vertical, downward).

Because the string is inextensible, the changes in the two segments satisfy

$d(L₁)/dt + d(L₂)/dt = 0$.

At the moment shown (position II) the string between the pulley and the sleeve makes an angle 53° with the horizontal. (The vertical distance from the pulley to the rod is given as 1.5 m.) The sleeve moves horizontally with speed $v$ but only the component along the string shortens $L₁$; that component is

$v₁ = v \cdot \cos53$.

Thus, $dL₁/dt = –v \cdot \cos53$ (negative because $L₁$ decreases).

Since the other segment $L₂$ (vertical) increases as the block falls, $dL₂/dt = v_M$ (taking downward as positive).

From (1) we have $–v \cdot \cos53 + v_M = 0 \implies v_M = v \cdot \cos53$.

Now, assume that when the system is set free the sleeve is at position I (say, exactly vertically below the pulley). When it reaches II the sleeve has moved a horizontal distance $d$ given by the geometry of the right‐triangle having vertical side 1.5 m and the string making 53° with the horizontal. In that triangle the hypotenuse (length $L₁$) is

$L₁ = (1.5)/\sin53$,

and the horizontal distance (sleeve displacement from the “vertical” point B) is

$d = L₁ \cdot \cos53 = (1.5 \cdot \cos53)/\sin53$.

Using numerical values ($\sin53 \approx 0.7986$, $\cos53 \approx 0.6018$): $d \approx (1.5 \times 0.6018)/0.7986 \approx 1.13$ m.

As the sleeve moves by this $d$, the vertical segment increases by exactly $d$ (due to the inextensibility and the geometry) so the block falls a distance

$\Delta y = d \cdot (\text{component along vertical}) = d \cdot \cos53$ (since the rate from constraint (2) gives $v_M = v \cos53$).

Thus, the loss in gravitational potential energy of the block is

$\Delta PE = M g (d \cdot \cos53)$.

This loss is converted into kinetic energies of the sleeve and the falling block: $K_{\text{total}} = (1/2)m v² + (1/2) M v_M²$ Since from (2) $v_M = v \cdot \cos53$, $K_{\text{total}} = (1/2) m v² + (1/2) M (v \cdot \cos53)² = (1/2) [m + M \cos²53] v²$.

Setting loss in potential energy equal to gain in kinetic energy, $M g (d \cdot \cos53) = (1/2)[m + M \cos²53] v²$.

Solve for $v²$: $v² = [2 M g d \cos53] / [m + M \cos²53]$.

Now substitute the numbers: $M = 25$ kg, $m = 9$ kg, $g = 9.81$ m/s², $\cos53 \approx 0.6018$, $\cos²53 \approx 0.3622$, $d \approx 1.13$ m.

Numerator: $2 \times 25 \times 9.81 \times 1.13 \times 0.6018 \approx 2 \times 25 = 50$, $50 \times 9.81 = 490.5$, $490.5 \times 1.13 \approx 554.07$, $554.07 \times 0.6018 \approx 333.13$.

Denominator: $9 + 25 \times 0.3622 = 9 + 9.055 = 18.055$. Thus, $v² \approx 333.13/18.055 \approx 18.45 \implies v \approx \sqrt{18.45} \approx 4.29$ m/s.

Rounded to the nearest integer, the sleeve’s speed is approximately 4 m/s.

Explanation of the solution (minimal):

1. Write the inextensibility constraint: $v_M = v \cdot \cos53$.
2. Determine the horizontal displacement $d$ using geometry: $d = (1.5 \cdot \cos53)/\sin53 \approx 1.13$ m.
3. Equate loss in gravitational potential energy (of block: $M g (d \cdot \cos53)$) to the total kinetic energy $[\frac{1}{2}(m + M \cos²53) v²]$.
4. Solve for $v$ to get $v \approx 4.29$ m/s, i.e. approximately 4 m/s.