Laws of Motion — Practice

Question 1

A book rests on a table. Name the third-law partner of the book's weight.

Accepted Answer

**The gravitational pull of the book on the Earth.**

The table's normal force is *not* the partner — it acts on the same body as the weight, and two forces on the same body can never form a third-law pair.

They happen to be equal here only because the book is in equilibrium. Put it in an accelerating lift and the normal force changes while the weight does not.

Question 2

A block on an incline has $\mu_s=0.6$. Find the angle of repose.

Accepted Answer

At the point of slipping, $mg\sin	heta=\mu_s mg\cos	heta$.

The mass and $g$ cancel: $	an	heta=\mu_s=0.6$

$	heta\approx\mathbf{31°}$

Note what cancelled — the angle depends only on the two surfaces, not on how heavy the block is.

Question 3

A lift accelerates upward at $2\ 	ext{m/s}^2$, $g=10$. Give the apparent weight of a 50 kg passenger.

Accepted Answer

Ground frame, up positive: $N-mg=ma$

$N=m(g+a)=50(12)=\mathbf{600\ 	ext{N}}$

True weight is 500 N. The extra 100 N is what accelerates the passenger upward — and it is what a bathroom scale in the lift would read.

Question 4

Masses 3 kg and 5 kg hang over a frictionless pulley, $g=10$. Find the acceleration and the tension.

Accepted Answer

5 kg: $50-T=5a$

3 kg: $T-30=3a$

Adding cancels $T$: $20=8a\Rightarrow a=\mathbf{2.5\ 	ext{m/s}^2}$

$T=30+3(2.5)=\mathbf{37.5\ 	ext{N}}$

*Check:* the tension lies between the two weights (30 N and 50 N), as it always must when one side descends.

Question 5

A 10 kg block has $\mu_s=0.5$, $\mu_k=0.4$ on a horizontal surface, $g=10$. A 40 N horizontal force is applied. Find the friction force and the acceleration.

Accepted Answer

Maximum static friction $=\mu_s N=0.5(100)=50\ 	ext{N}$.

The applied 40 N is **less** than 50 N, so the block does not move.

Friction $=\mathbf{40\ 	ext{N}}$ exactly — whatever is needed to hold it — and $a=\mathbf{0}$.

Only above 50 N does it slide, and the instant it does, friction *drops* to $\mu_k N=40\ 	ext{N}$. That sudden drop is why objects lurch when they start moving.

Question 6

Blocks of 2 kg and 4 kg are in contact on a frictionless surface. A 12 N force pushes the 2 kg block. Find the acceleration and the contact force, then repeat with the force applied to the 4 kg block.

Accepted Answer

Whole system: $a=\dfrac{12}{6}=\mathbf{2\ 	ext{m/s}^2}$ in both cases — the acceleration cannot depend on which end you push.

**Pushing the 2 kg side:** the contact force accelerates the 4 kg block, so $F_c=4(2)=\mathbf{8\ 	ext{N}}$.

**Pushing the 4 kg side:** it accelerates the 2 kg block, so $F_c=2(2)=\mathbf{4\ 	ext{N}}$.

Same acceleration, different internal force. The contact force is only responsible for what is in front of it.

Question 7

A curve of radius $100\ 	ext{m}$ is banked at $30°$. Find the design speed, $g=10$.

Accepted Answer

Frictionless banking: $N\sin	heta=\dfrac{mv^{2}}{r}$ and $N\cos	heta=mg$.

Dividing: $	an	heta=\dfrac{v^{2}}{rg}$

$v^{2}=rg	an 30°=1000(0.577)=577\Rightarrow v\approx\mathbf{24\ 	ext{m/s}}$

Both $m$ and $N$ cancelled, so this design speed is the same for a motorcycle and a loaded truck.

Question 8

For that same banked curve with $\mu=0.2$, find the maximum safe speed and explain why friction raises it.

Accepted Answer

$v_{max}=\sqrt{rg\dfrac{\mu+	an	heta}{1-\mu	an	heta}}=\sqrt{1000	imes\dfrac{0.777}{0.885}}\approx\mathbf{29.6\ 	ext{m/s}}$

Above the design speed the vehicle tends to slide **outward and up** the bank, so friction acts **inward and down** the slope — adding to the horizontal component of $N$ rather than fighting it.

There is a matching $v_{min}$ below which it slides inward and friction reverses.

Question 9

A block rests on an incline of angle $	heta$ with friction coefficient $\mu$, where $	an	heta>\mu$. It is held by a horizontal force $F$. Derive the minimum $F$.

Accepted Answer

$	an	heta>\mu$ means the block would slide down unaided, so friction acts **up** the slope and $F$ must supply the rest.

Perpendicular to the incline: $N=mg\cos	heta+F\sin	heta$ — note $F$ presses the block in, increasing $N$.

Along the incline, at the verge of sliding down:

$mg\sin	heta=F\cos	heta+\mu N$

Substituting and solving:

$F_{min}=mg\dfrac{\sin	heta-\mu\cos	heta}{\cos	heta+\mu\sin	heta}$

*Check:* if $	an	heta=\mu$ the numerator vanishes and $F_{min}=0$ — exactly the angle of repose, where no force is needed.

Question 10

A pendulum hangs in a train accelerating at $5\ 	ext{m/s}^2$, $g=10$. Find its angle to the vertical, from both the ground frame and the train frame.

Accepted Answer

**Ground frame.** The bob accelerates forward, so the net force is horizontal and forward:

$T\sin	heta=ma$, $T\cos	heta=mg$

$	an	heta=\dfrac{a}{g}=0.5\Rightarrow	heta\approx\mathbf{26.6°}$

**Train frame.** Add a backward pseudo force $ma$. The bob is now in equilibrium under three forces, and resolving gives the identical equations.

Same angle. Different story, same prediction — which is the point of pseudo forces.

Question 11

A 5 kg shell moving at $20\ 	ext{m/s}$ explodes into two equal fragments. One is left at rest. Find the velocity of the other, and comment on the kinetic energy.

Accepted Answer

Explosive forces are internal, so total momentum is unchanged:

$5(20)=2.5(0)+2.5v \Rightarrow v=\mathbf{40\ 	ext{m/s}}$, along the original direction.

$K_i=	frac12(5)(20)^{2}=1000\ 	ext{J}$

$K_f=	frac12(2.5)(40)^{2}=2000\ 	ext{J}$

**The kinetic energy doubled.** That is not a violation — the extra 1000 J came from the chemical energy of the explosive. Momentum conservation constrains an explosion; energy conservation does not, because energy was stored in the system beforehand.

Question 12

A 4 kg block on a rough table ($\mu_k=0.25$) is joined by a string over a frictionless pulley to a 2 kg block hanging freely. $g=10$. Find the acceleration and the tension.

Accepted Answer

Friction on the table block: $f=\mu_k m_1 g=0.25(4)(10)=10\ 	ext{N}$.

Hanging block: $m_2g-T=m_2a \Rightarrow 20-T=2a$

Table block: $T-f=m_1a \Rightarrow T-10=4a$

Adding removes $T$: $10=6a \Rightarrow a=\mathbf{1.67\ 	ext{m/s}^{2}}$

$T=20-2(1.67)=\mathbf{16.7\ 	ext{N}}$

*Check:* substituting into the table block's equation gives $16.7-10=6.7=4(1.67)$. Consistent.

*Sanity check:* $T$ is less than the hanging weight of 20 N, as it must be for that block to accelerate downward. Had friction exceeded 20 N, nothing would have moved at all and $T$ would have been 20 N.

Practice — Laws of Motion

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