2024 A/L Physics Past Paper · Question 16 (English Medium)

Answer: (1) $0{\cdot}7$

Given information

This question is about pushing two boxes that touch, along a rough floor.

A force of $100\ \text{N}$ pushes box $A$. Box $A$ pushes box $B$ in front of it, so both slide together.

The boxes are made of different materials, so the floor rubs on each one differently. We know the friction for $A$. We also know that $A$ and $B$ push on each other with $40\ \text{N}$.

What we must find: the coefficient of kinetic friction between box $B$ and the floor.

How we find it: use $A$ alone to find the acceleration they share. Then use $B$ alone to find its friction.

$m_A=10\ \text{kg}$ and $m_B=5\ \text{kg}$. The push on $A$ is $F=100\ \text{N}$. For $A$, $\mu_A=0.5$. The force between the boxes is $P=40\ \text{N}$. Take $g=10\ \text{m s}^{-2}$.

What the question asks

$m_A$, $m_B$ ($\text{kg}$): masses of $A$ and $B$.

$F$ ($\text{N}$): horizontal force on $A$.

$P$ ($\text{N}$): reaction force between $A$ and $B$.

$R_A$, $R_B$ ($\text{N}$): normal contact forces from the floor.

$F_A$, $F_B$ ($\text{N}$): kinetic friction forces on $A$ and $B$.

$\mu_A$, $\mu_B$: coefficients of kinetic friction.

$a$ ($\text{m s}^{-2}$): acceleration of both boxes.

Relevant theory

$\therefore F = ma$

F - Resultant force in newtons m - mass in kg a - acceleration in m s^-2

The dynamic friction acting on an object is also proportional to the normal contact force between surfaces. If the dynamic friction is $F_d$ and normal contact force is R,

$F_d \propto R$ $F_d = \mu_d R$

$\mu_d$ is called the coefficient of dynamic friction.

When the objects interact (collide), they exert equal and oppsite forces on one another.

What is happening: the boxes stay in contact, so they speed up together with the same $a$.

Box $A$ is pushed forward by $100\ \text{N}$. It is held back by its friction and by $B$, which pushes back on it with $40\ \text{N}$.

Box $B$ is pushed forward only by $A$, with the same $40\ \text{N}$. It is held back only by its own friction.

The floor is horizontal, so each normal contact force equals the weight: $R=mg$.

For example, push two school desks in a line. You push only the back desk.

The back desk pushes the front desk. That push is the only thing moving the front desk forward.

If the front desk drags hard on the floor, it needs a bigger push to keep up.

In words: look at each desk on its own. The push between them tells us how hard the floor drags on the front one.

Required equations

$F_A=\mu_AR_A=\mu_Am_Ag$

$F-F_A-P=m_Aa$

$P-\mu_Bm_Bg=m_Ba$

Applying it to this question

$100$

$R_A=10\times10$

$50$

$F_A=0.5\times100$

$100-50-40=10a$

$10=10a$

$1$

$a$

$40-F_B=5\times1$

$35$

$F_B$

$R_B=5\times10$

$\mu_B=\dfrac{35}{50}$

$0.7$

$\mu_B$

Reasoning and calculation

Both boxes must have $a=1\ \text{m s}^{-2}$. Only $\mu_B=0.7$ leaves $B$ with exactly $40-35=5\ \text{N}$ to give its $5\ \text{kg}$ that acceleration. This is option (1).

With $0.6$, the friction on $B$ is $30\ \text{N}$. Then $B$ would have $a=(40-30)/5=2\ \text{m s}^{-2}$. But $A$ has $a=1\ \text{m s}^{-2}$, and the boxes move together.

$0.5$ copies the coefficient of $A$. The boxes are made of different materials, so there is no reason for the same value. With $0.5$, $B$ would have $a=(40-25)/5=3\ \text{m s}^{-2}$, not $1\ \text{m s}^{-2}$.

$0.4$ comes from $40/100$, the push between the boxes divided by the applied force. Friction coefficient is friction divided by $R$. With $0.4$, $B$ would have $a=(40-20)/5=4\ \text{m s}^{-2}$.

With $0.3$, the friction on $B$ is only $15\ \text{N}$. Then $B$ would have $a=(40-15)/5=5\ \text{m s}^{-2}$, five times the acceleration of $A$. $B$ would run away from $A$, but the $40\ \text{N}$ push says they stay in contact.

Checks

$(40-5*1)/(5*10)$

$F-F_A-F_B=(m_A+m_B)a$

$100-50-35=15\times1$

$15\ \text{N}=15\ \text{N}$

$\mu_B=F_B/R_B$ is $\text{N}$ divided by $\text{N}$, so it has no unit, as a friction coefficient should.

Taken together, the inside pushes of $40\ \text{N}$ cancel, and the check works without them. The friction on $B$, $35\ \text{N}$, is less than the $40\ \text{N}$ push it gets, so $B$ speeds up forwards, as it should.

Conclusion

$\mu_B=0.7$

The coefficient of kinetic friction for $B$ is $0.7$. The answer is (1).