2020 A/L Physics Past Paper · Question 22 (English Medium)

Blocks joined over a pulley — force for constant speed

Two blocks $A$ , $B$ of mass $0.5\ \text{kg}$ , $1.0\ \text{kg}$ respectively are connected as shown in the figure by a light inextensible string passing over a light smooth pulley. The coefficient of dynamic friction between every pair of surfaces in contact is $0.25$ . What is the force $F$ that must be applied on block $B$ to pull it towards the left with constant speed?

(1) $2.50$ N
(2) $3.75$ N
(3) $5.00$ N
(4) $6.25$ N
(5) $7.50$ N

Answer and explanation

Answer: (4) $6.25$ N

Given information

We know: $m_A = 0.5$ kg, $m_B = 1.0$ kg.

Between every pair of surfaces in contact, $\mu = 0.25$ .

$g = 10\ \text{m s}^{-2}$ .

The pulley is light and smooth.

B moves towards the left at constant speed.

What the question asks

We must find: the force $F$ needed to pull B at constant speed.

Relevant theory

Every object continues to be at rest or move with uniform velocity unless acted upon by a resultant force.

The friction acting between two surfaces as they slide over one another is called dynamic friction.

$F_d = \mu_d R$

Required equations

$f = \mu R$

$T - f_1 = 0$

$F - f_1 - f_2 - T = 0$

Applying it to this question

Forces on block A: weight $5$ N (down), normal reaction from B $5$ N (up).

The tension $T$ acts on A towards the right, towards the pulley.

The friction $f_1$ on B's surface acts on A towards the left — since A slides to the right.

Forces on block B: the applied force $F$ towards the left.

A's friction $f_1$ acts on B towards the right, the floor's friction $f_2$ towards the right, and the tension $T$ towards the right.

A presses down on B with $5$ N, and B's own weight is $10$ N; so the floor's normal reaction is $15$ N.

Reasoning and calculation

Equilibrium condition: both blocks move at constant speed.

By Newton's first law, the resultant force on each one is zero.

$1.25$

$f_1 = 0.25 \times 0.5 \times 10$

$T = f_1$

$15$

$R_2 = (0.5 + 1.0) \times 10$

$3.75$

$f_2 = 0.25 \times 15$

$6.25$

$F = 1.25 + 3.75 + 1.25$

Checks

Second way: treat A, B and the string as one system.

Now $f_1$ is an internal force and cancels out.

The remaining external horizontal forces are: $F$ (left), the floor's friction $f_2$ (right), and $2T$ (right) from the wall through the pulley's axle.

$F = f_2 + 2T = 3.75 + 2 \times 1.25 = 6.25$ N ✓.

$2 * mu * mA * g + mu * (mA + mB) * g$

Conclusion

$F = 6.25\ \text{N}$

$F = f_1 + f_2 + T = 1.25 + 3.75 + 1.25 = 6.25$ N — the answer is (4).

Forgetting the tension acting on B.

One end of the string is on A and the other on B: it pulls both blocks towards the pulley with $T$ .

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