601. A \(5\,\text{kg}\) block is pulled horizontally on a rough horizontal surface. The applied force \(P\) is increased slowly from zero. The maximum static friction is \(20\,\text{N}\), while the kinetic friction after sliding begins is \(15\,\text{N}\). Take \(g=10\,\text{m s}^{-2}\). Which description of the block's response is correct?
ⓐ. For \(0\le P\lt 20\,\text{N}\), \(f=P\); at \(P=20\,\text{N}\), the friction must already be kinetic.
ⓑ. For \(0\lt P\lt 20\,\text{N}\), \(f=20\,\text{N}\); just above \(20\,\text{N}\), \(f=15\,\text{N}\) and the block remains at rest.
ⓒ. For \(0\lt P\lt 15\,\text{N}\), \(f=P\); at \(P=15\,\text{N}\), kinetic friction starts and the block begins moving.
ⓓ. For \(0\le P\le20\,\text{N}\), \(f=P\); just above \(20\,\text{N}\), \(f=15\,\text{N}\) and the block has non-zero acceleration.
Correct Answer: For \(0\le P\le20\,\text{N}\), \(f=P\); just above \(20\,\text{N}\), \(f=15\,\text{N}\) and the block has non-zero acceleration.
Explanation: Before sliding begins, static friction is self-adjusting. Therefore,
\[
f_s=P
\]
as long as the required value does not exceed the limiting value \(20\,\text{N}\). The relation is not \(f_s=20\,\text{N}\) throughout the static regime. At
\[
P=20\,\text{N},
\]
the block is at limiting equilibrium and can still be instantaneously at rest. When \(P\) is increased beyond this threshold, static friction can no longer maintain equilibrium. Relative sliding then begins and the appropriate model changes to kinetic friction. The friction magnitude becomes \(15\,\text{N}\), according to the stated data. Hence the applied force immediately exceeds the opposing kinetic friction, producing a non-zero resultant and acceleration. The smaller kinetic value does not determine the onset of motion; the onset is governed by the limiting static friction.
602. A force sensor is used to study the same pair of dry surfaces under four trials.
| Trial | Normal reaction \(N\) | Regime | Measured friction |
|---|
| P | \(50\,\text{N}\) | Impending motion | \(20\,\text{N}\) |
| Q | \(80\,\text{N}\) | Impending motion | \(32\,\text{N}\) |
| R | \(50\,\text{N}\) | Sliding | \(15\,\text{N}\) |
| S | \(80\,\text{N}\) | Sliding | \(24\,\text{N}\) |
Which conclusion is best supported?
ⓐ. \(\mu_s=0.40,\ \mu_k=0.30\), and both limiting and kinetic friction are proportional to \(N\) within the elementary model.
ⓑ. \(\mu_s=0.30,\ \mu_k=0.40\), because sliding friction must exceed static friction.
ⓒ. The coefficient of friction has units of newtons.
ⓓ. Trial Q is inconsistent because its friction differs from Trial P.
Correct Answer: \(\mu_s=0.40,\ \mu_k=0.30\), and both limiting and kinetic friction are proportional to \(N\) within the elementary model.
Explanation: From Trial P,
\[
\mu_s=\frac{20}{50}=0.40.
\]
Trial Q gives
\[
\mu_s=\frac{32}{80}=0.40,
\]
confirming the same limiting-friction ratio at a different normal reaction. From Trial R,
\[
\mu_k=\frac{15}{50}=0.30.
\]
Trial S independently gives
\[
\mu_k=\frac{24}{80}=0.30.
\]
Thus the data are internally consistent with
\[
f_{s,\max}=\mu_sN
\]
and
\[
f_k=\mu_kN.
\]
The coefficients are ratios of forces and are therefore dimensionless. The larger friction in Q than in P is expected because \(N\) is larger, not because the surface pair has changed.
603. A \(4\,\text{kg}\) block is initially at rest on a horizontal rough surface. The coefficients are \(\mu_s=0.50\) and \(\mu_k=0.30\). A constant horizontal force of \(16\,\text{N}\) acts on the block. An upward vertical force \(Y\) is then slowly increased from zero while the horizontal force remains unchanged. Take \(g=10\,\text{m s}^{-2}\). At what value of \(Y\) does sliding begin, and what is the horizontal acceleration immediately after sliding begins at that same value of \(Y\)?
ⓐ. \(Y=8\,\text{N}\); \(a=0\)
ⓑ. \(Y=16\,\text{N}\); \(a=1.0\,\text{m s}^{-2}\)
ⓒ. \(Y=8\,\text{N}\); \(a=1.6\,\text{m s}^{-2}\)
ⓓ. \(Y=20\,\text{N}\); \(a=2.5\,\text{m s}^{-2}\)
Correct Answer: \(Y=8\,\text{N}\); \(a=1.6\,\text{m s}^{-2}\)
Explanation: The upward force reduces the normal reaction:
\[
N=40-Y.
\]
The block remains at rest while the required static friction \(16\,\text{N}\) does not exceed its limiting value. At the threshold,
\[
16=\mu_sN.
\]
Therefore,
\[
16=0.50(40-Y).
\]
Hence,
\[
40-Y=32,
\]
so
\[
Y=8\,\text{N}.
\]
Once sliding begins, the friction model changes to
\[
f_k=\mu_kN=0.30(32)=9.6\,\text{N}.
\]
The horizontal resultant is
\[
16-9.6=6.4\,\text{N},
\]
giving
\[
a=\frac{6.4}{4}=1.6\,\text{m s}^{-2}.
\]
The question therefore requires first finding the static threshold and only then switching to the kinetic-friction relation.
604. A rectangular block is tested on the same rough horizontal surface in two orientations. In orientation P, its apparent contact area is \(A\); in orientation Q, it is \(2A\). In both orientations the normal reaction is \(60\,\text{N}\). The measured limiting friction values are \(24.1\,\text{N}\) and \(23.9\,\text{N}\), respectively. Which conclusion is most appropriate?
ⓐ. The limiting friction is inversely proportional to area; the two readings imply different \(\mu_s\) values because the areas differ.
ⓑ. The limiting friction is approximately area-independent; both readings give \(\mu_s\approx0.40\) within the measurement variation.
ⓒ. Doubling the apparent area halves \(\mu_s\); the nearly equal limiting forces therefore contradict the elementary model.
ⓓ. The elementary model is invalid because two trials with the same \(N\) must give exactly identical limiting-friction readings.
Correct Answer: The limiting friction is approximately area-independent; both readings give \(\mu_s\approx0.40\) within the measurement variation.
Explanation: The two measurements are very close even though the apparent contact area is doubled. Their corresponding ratios are
\[
\frac{24.1}{60}\approx0.402
\]
and
\[
\frac{23.9}{60}\approx0.398.
\]
Both are consistent with
\[
\mu_s\approx0.40.
\]
The elementary dry-friction law treats limiting friction as approximately independent of apparent contact area when the normal reaction and surface pair are otherwise unchanged. Experimental friction laws are not expected to behave as mathematically exact identities under all real conditions. Therefore a small difference such as \(24.1\) versus \(23.9\,\text{N}\) is not evidence of an area dependence. The controlled feature of this experiment is that \(N\) and the material pair remain unchanged. Hence option B is the scientifically appropriate interpretation.
605. A \(5\,\text{kg}\) block rests on a rough horizontal surface with coefficient of static friction \(\mu_s=0.40\). Two horizontal forces act simultaneously on it: \(12\,\text{N}\) eastward and \(16\,\text{N}\) northward. Later, only the eastward force is increased from \(12\,\text{N}\) to \(15\,\text{N}\), while the northward force remains \(16\,\text{N}\). Take \(g=10\,\text{m s}^{-2}\). Which statement is correct?
ⓐ. Initially the friction is \(16\,\text{N}\) southward because static friction opposes only the larger applied-force component.
ⓑ. Initially \(\vec f_s=(-12\hat i-16\hat j)\,\text{N}\); after the change, static friction becomes \((-15\hat i-16\hat j)\,\text{N}\) and the block still remains at rest.
ⓒ. Initially \(\vec f_s=(-12\hat i-16\hat j)\,\text{N}\) and the block is at limiting equilibrium; after the change, the block cannot remain at rest because the required friction would have magnitude \(\sqrt{481}\,\text{N}\gt 20\,\text{N}\).
ⓓ. Initially the friction is \(20\,\text{N}\) westward; after the change the limiting friction automatically increases to \(\sqrt{481}\,\text{N}\).
Correct Answer: Initially \(\vec f_s=(-12\hat i-16\hat j)\,\text{N}\) and the block is at limiting equilibrium; after the change, the block cannot remain at rest because the required friction would have magnitude \(\sqrt{481}\,\text{N}\gt 20\,\text{N}\).
Explanation: The vertical forces balance, so the normal reaction is
\[
N=mg=5(10)=50\,\text{N}.
\]
Hence the maximum available static friction is
\[
f_{s,\max}=\mu_sN=0.40(50)=20\,\text{N}.
\]
Initially, the resultant of the two applied horizontal forces is
\[
\vec F_{\text{app}}=12\hat i+16\hat j\,\text{N}.
\]
Its magnitude is
\[
|\vec F_{\text{app}}|=\sqrt{12^2+16^2}=20\,\text{N}.
\]
For equilibrium, static friction must be equal and opposite:
\[
\vec f_s=-12\hat i-16\hat j\,\text{N}.
\]
Its magnitude is exactly \(20\,\text{N}\), so the block is at limiting equilibrium. After the eastward component is increased to \(15\,\text{N}\), equilibrium would require friction of magnitude
\[
\sqrt{15^2+16^2}=\sqrt{481}\,\text{N}\approx21.9\,\text{N}.
\]
This exceeds \(f_{s,\max}=20\,\text{N}\). Therefore static friction can no longer maintain rest, and the block must begin to slip. Static friction is a vector that opposes the net tendency of relative motion; its components are not treated independently as separate friction limits.
606. The kinetic friction measured for the same pair of dry surfaces is recorded at different sliding speeds while the normal reaction is held fixed at \(100\,\text{N}\).
| Speed (\(\text{m s}^{-1}\)) | 0.5 | 1.0 | 2.0 | 4.0 |
|---|
| Measured friction (N) | 29.8 | 30.1 | 30.0 | 30.2 |
Which interpretation is most consistent with the elementary friction model?
ⓐ. \(\mu_k\approx0.30\), and the measurements support approximate near-independence from moderate sliding-speed changes.
ⓑ. Friction is exactly proportional to speed because the largest speed gives the largest measured friction.
ⓒ. \(\mu_k\) changes from \(0.298\) to \(0.302\), proving that no coefficient can be assigned.
ⓓ. Kinetic friction must decrease with speed, so the data are impossible.
Correct Answer: \(\mu_k\approx0.30\), and the measurements support approximate near-independence from moderate sliding-speed changes.
Explanation: Since the normal reaction is fixed,
\[
\mu_k=\frac{f_k}{N}.
\]
The measured values correspond to ratios clustered very near
\[
0.30.
\]
For example,
\[
\frac{30.0}{100}=0.30.
\]
The small scatter from \(29.8\) to \(30.2\,\text{N}\) is tiny compared with the overall friction magnitude. The elementary dry-friction model therefore treats the kinetic friction as nearly independent of moderate sliding-speed changes over this tested range. It does not claim universal exact speed independence under every physical condition. The experiment is well designed because speed is varied while the normal reaction and surface pair are held fixed. Hence option A correctly combines the numerical evidence with the empirical qualifier.
607. Consider the following statements.
Statement I: The coefficient of friction is dimensionless.
Statement II: For ordinary dry surfaces in the elementary model, \(\mu_k\) is commonly smaller than \(\mu_s\).
Statement III: The actual static friction on a body at rest must always equal \(\mu_sN\).
Statement IV: Kinetic friction is modeled as approximately proportional to the normal reaction.
Which option is correct?
ⓐ. I, II, III and IV
ⓑ. I and III only
ⓒ. II, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: A coefficient of friction is a ratio of two forces, such as
\[
\mu_s=\frac{f_{s,\max}}{N}.
\]
Therefore it is dimensionless, making Statement I correct. In the standard elementary treatment of ordinary dry surfaces,
\[
\mu_k\lt \mu_s
\]
is a common relation, so Statement II is correct. Statement III is false because static friction is self-adjusting:
\[
0\le f_s\le\mu_sN.
\]
Equality holds only at limiting equilibrium. For actual sliding, the elementary law is
\[
f_k\approx\mu_kN,
\]
so Statement IV is correct. The distinction between an inequality for static friction and an approximate equality for kinetic friction is essential.
608. A \(5\,\text{kg}\) block rests on a horizontal rough surface. A force sensor records the friction while a horizontal applied force \(P\) is increased gradually.
From \(P=0\) to \(P=25\,\text{N}\), the measured friction magnitude rises along the line \(f=P\). At \(P=25\,\text{N}\), motion begins. Immediately after motion starts, the measured friction drops to \(20\,\text{N}\), and while sliding it remains approximately \(20\,\text{N}\) over the measured force range.
Take \(g=10\,\text{m s}^{-2}\). Which option is correct?
ⓐ. \(\mu_s=\mu_k=0.50\); at \(P=35\,\text{N}\), \(a=2\,\text{m s}^{-2}\)
ⓑ. \(\mu_s=0.40,\ \mu_k=0.50\); at \(P=35\,\text{N}\), \(a=3\,\text{m s}^{-2}\)
ⓒ. \(\mu_s=0.50,\ \mu_k=0.40\); at \(P=35\,\text{N}\), \(a=7\,\text{m s}^{-2}\)
ⓓ. \(\mu_s=0.50,\ \mu_k=0.40\); at \(P=35\,\text{N}\), \(a=3\,\text{m s}^{-2}\)
Correct Answer: \(\mu_s=0.50,\ \mu_k=0.40\); at \(P=35\,\text{N}\), \(a=3\,\text{m s}^{-2}\)
Explanation: The normal reaction is
\[
N=mg=50\,\text{N}.
\]
The maximum static friction read from the graph is
\[
f_{s,\max}=25\,\text{N}.
\]
Hence,
\[
\mu_s=\frac{25}{50}=0.50.
\]
The kinetic friction after sliding begins is
\[
f_k=20\,\text{N},
\]
so
\[
\mu_k=\frac{20}{50}=0.40.
\]
At \(P=35\,\text{N}\), the block is already sliding. The resultant horizontal force is
\[
35-20=15\,\text{N}.
\]
Therefore,
\[
a=\frac{15}{5}=3\,\text{m s}^{-2}.
\]
The drop at the onset of sliding is consistent with the ordinary relation \(\mu_k\lt \mu_s\).
609. A \(10\,\text{kg}\) crate is pulled horizontally by a force that is slowly increased. Initially its normal reaction is \(100\,\text{N}\), and sliding begins at \(40\,\text{N}\). Without changing the contacting materials, an additional downward force increases the normal reaction to \(150\,\text{N}\). What horizontal force is now required to reach limiting equilibrium, and what has happened to the coefficient of static friction?
ⓐ. \(90\,\text{N}\); \(\mu_s\) remains \(0.60\)
ⓑ. \(40\,\text{N}\); \(\mu_s\) becomes \(0.60\)
ⓒ. \(60\,\text{N}\); \(\mu_s\) remains \(0.40\)
ⓓ. \(60\,\text{N}\); \(\mu_s\) becomes \(0.60\)
Correct Answer: \(60\,\text{N}\); \(\mu_s\) remains \(0.40\)
Explanation: The original threshold gives
\[
\mu_s=\frac{40}{100}=0.40.
\]
The contacting materials are unchanged, so in the elementary model the coefficient remains associated with the same surface pair. The additional downward force changes the normal reaction, not the coefficient itself. With
\[
N'=150\,\text{N},
\]
the new limiting static friction is
\[
f_{s,\max}'=\mu_sN'.
\]
Thus,
\[
f_{s,\max}'=0.40(150)=60\,\text{N}.
\]
Therefore the applied horizontal force must reach \(60\,\text{N}\) for impending motion. The larger frictional threshold arises from the larger normal reaction, not from a change in \(\mu_s\).
610. Two identical blocks are made from the same material. Block P rests on surface X and has limiting friction \(24\,\text{N}\) for a normal reaction of \(60\,\text{N}\). Block Q rests on surface Y and has limiting friction \(30\,\text{N}\) for a normal reaction of \(100\,\text{N}\). Which statement is correct?
ⓐ. The coefficients are equal because the blocks are identical.
ⓑ. Surface pair Q has the larger coefficient because \(30\,\text{N}\gt 24\,\text{N}\).
ⓒ. Surface pair P has the larger coefficient of static friction even though its limiting friction force is smaller.
ⓓ. Coefficients cannot be compared unless the apparent contact areas are known.
Correct Answer: Surface pair P has the larger coefficient of static friction even though its limiting friction force is smaller.
Explanation: For P,
\[
\mu_{s,P}=\frac{24}{60}=0.40.
\]
For Q,
\[
\mu_{s,Q}=\frac{30}{100}=0.30.
\]
Thus,
\[
\mu_{s,P}\gt \mu_{s,Q}.
\]
The raw friction-force magnitudes alone do not determine which surface pair has the larger coefficient because the normal reactions are different. A larger \(N\) can produce a larger limiting friction even with a smaller coefficient. The identity of the block material alone also does not fix the coefficient because friction depends on the interacting pair of surfaces. Apparent contact area is not required for this elementary comparison. Therefore P has the larger coefficient despite its smaller measured limiting force.
611. A car wheel rolls without slipping while the car accelerates forward on a level road. Later, the same wheel is locked during hard braking and slides over the road. Which statement correctly identifies the tyre-road friction regimes?
ⓐ. The locked wheel experiences no friction because it no longer rotates.
ⓑ. Both cases must involve kinetic friction because the car moves relative to the road.
ⓒ. Both cases must involve static friction because the road itself is stationary.
ⓓ. The accelerating rolling wheel can involve static friction, while the locked sliding wheel involves kinetic friction.
Correct Answer: The accelerating rolling wheel can involve static friction, while the locked sliding wheel involves kinetic friction.
Explanation: The relevant criterion is relative motion at the contact, not whether the vehicle as a whole moves relative to the ground. During rolling without slipping, the tyre element in contact with the road is instantaneously at rest relative to the road. The frictional interaction at that contact is therefore static. That static friction can be essential for traction and forward acceleration. When the wheel is locked and slides over the road, relative sliding exists at the contact. The appropriate regime is then kinetic friction. Stopping the wheel from rotating does not eliminate friction; it changes the nature of the contact motion. Thus friction may be useful in both situations, but the physical regime is different.
612. A machine contains two contacts. Contact P is a dry sliding guide producing unwanted heating. Contact Q is a driven wheel that must roll without slipping to transmit a horizontal force to the ground. Which modification is most appropriate?
ⓐ. Reduce friction strongly at P by lubrication while preserving adequate static grip at Q.
ⓑ. Reduce friction strongly at both P and Q because friction always wastes energy.
ⓒ. Increase friction strongly at P but eliminate it at Q.
ⓓ. Replace the wheel at Q by a freely sliding block because kinetic friction is always larger than static friction.
Correct Answer: Reduce friction strongly at P by lubrication while preserving adequate static grip at Q.
Explanation: In the sliding guide, kinetic friction acts as an unwanted resistance. It can produce heating, wear and power loss. Lubrication is therefore useful because it can reduce this undesirable sliding resistance. At the driven wheel, however, traction without slipping relies on static friction at the tyre-ground contact. If the available static friction is reduced too much, the wheel may slip instead of transmitting the required force. Thus friction is not intrinsically beneficial or harmful in every context. Its engineering role depends on what the contact is supposed to accomplish. A good design may deliberately reduce friction at one contact while preserving adequate friction at another. Option A correctly distinguishes those two functional roles.
613. Assertion: A block may remain at rest on a rough horizontal surface even when the maximum available static friction increases substantially.
Reason: Increasing the normal reaction increases \(\mu_sN\), but the actual static friction is determined by the force required for equilibrium and need not increase if that requirement is unchanged.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: Static friction is not automatically equal to its limiting value. Its elementary range is
\[
0\le f_s\le\mu_sN.
\]
Suppose a horizontal force of \(10\,\text{N}\) is being balanced by \(10\,\text{N}\) of static friction. If an additional downward force increases \(N\), the maximum possible static friction \(\mu_sN\) increases. However, the horizontal force that needs balancing is still only \(10\,\text{N}\). The actual friction therefore remains \(10\,\text{N}\), provided that value lies below the new limit. Hence increasing frictional capacity does not force the actual static friction to increase. The Reason directly explains the Assertion.
614. A \(5\,\text{kg}\) block is subjected to the following sequence on the same horizontal rough surface. The coefficients are \(\mu_s=0.40\) and \(\mu_k=0.30\), and \(g=10\,\text{m s}^{-2}\).
Stage I: a horizontal force is raised slowly from \(0\) to \(18\,\text{N}\).
Stage II: it is raised further to \(25\,\text{N}\).
Stage III: while the block is sliding, the force is reduced to \(12\,\text{N}\).
Which sequence of friction magnitudes is correct?
ⓐ. Stage I: \(15\,\text{N}\); Stage II: \(15\,\text{N}\); Stage III: \(12\,\text{N}\)
ⓑ. Stage I: \(20\,\text{N}\); Stage II: \(20\,\text{N}\); Stage III: \(12\,\text{N}\)
ⓒ. Stage I: \(18\,\text{N}\); Stage II: \(25\,\text{N}\); Stage III: \(12\,\text{N}\)
ⓓ. Stage I: \(18\,\text{N}\); Stage II: \(15\,\text{N}\); Stage III: \(15\,\text{N}\) as long as sliding continues
Correct Answer: Stage I: \(18\,\text{N}\); Stage II: \(15\,\text{N}\); Stage III: \(15\,\text{N}\) as long as sliding continues
Explanation: The normal reaction is
\[
N=50\,\text{N}.
\]
Therefore,
\[
f_{s,\max}=0.40(50)=20\,\text{N}
\]
and
\[
f_k=0.30(50)=15\,\text{N}.
\]
In Stage I, \(18\,\text{N}\) is below the static limit, so the block remains at rest and actual friction is \(18\,\text{N}\). In Stage II, the applied force exceeds \(20\,\text{N}\), so sliding begins and friction becomes \(15\,\text{N}\). In Stage III, the block is explicitly stated to still be sliding when the applied force is reduced. Therefore kinetic friction continues to have magnitude \(15\,\text{N}\), not \(12\,\text{N}\). The block may decelerate and later stop, but the friction regime cannot be changed merely because the applied force has become smaller than the kinetic-friction magnitude.
615. A small block is sliding on a horizontal plate. At an instant, the velocity of the block relative to the plate is
\[
\vec v_{\text{rel}}=(3\hat i-4\hat j)\,\text{m s}^{-1}.
\]
The normal reaction is \(50\,\text{N}\) and the coefficient of kinetic friction is \(0.30\). The plate itself may be translating relative to the ground, but the relative-velocity vector above is known. What is the kinetic-friction force exerted by the plate on the block?
ⓐ. \((-12\hat i+9\hat j)\,\text{N}\)
ⓑ. \((9\hat i-12\hat j)\,\text{N}\)
ⓒ. \(-15\hat i\,\text{N}\)
ⓓ. \((-9\hat i+12\hat j)\,\text{N}\)
Correct Answer: \((-9\hat i+12\hat j)\,\text{N}\)
Explanation: Kinetic friction opposes the relative sliding velocity at the contact, not necessarily the velocity of the block relative to the ground. Its magnitude is
\[
f_k=\mu_kN=0.30(50)=15\,\text{N}.
\]
The magnitude of the given relative velocity is
\[
|\vec v_{\text{rel}}|=\sqrt{3^2+(-4)^2}=5\,\text{m s}^{-1}.
\]
Therefore the unit vector along the relative sliding direction is
\[
\hat v_{\text{rel}}=\frac35\hat i-\frac45\hat j.
\]
The friction direction is opposite to this vector, so
\[
\vec f_k
=-15\hat v_{\text{rel}}
=-15\left(\frac35\hat i-\frac45\hat j\right).
\]
Hence,
\[
\vec f_k=(-9\hat i+12\hat j)\,\text{N}.
\]
Its magnitude is \(\sqrt{(-9)^2+12^2}=15\,\text{N}\), as required. Also,
\[
\vec f_k\cdot\vec v_{\text{rel}}=(-9)(3)+(12)(-4)=-75\lt 0,
\]
which confirms that friction opposes the relative sliding. The plate's separate motion relative to the ground does not alter this conclusion once \(\vec v_{\text{rel}}\) is specified.
616. A researcher proposes four conclusions after testing dry friction.
Statement I: Because \(\mu\) is dimensionless, it must have the same value for every pair of materials.
Statement II: A larger normal reaction can produce a larger friction force even when the coefficient is unchanged.
Statement III: A block with no tendency to slide can have zero static friction even though \(\mu_sN\) is non-zero.
Statement IV: Approximate area- and speed-independence statements should be applied only within the ordinary empirical regime for which they are intended.
Which option is correct?
ⓐ. II, III and IV only
ⓑ. I and II only
ⓒ. I, III and IV only
ⓓ. I, II, III and IV
Correct Answer: II, III and IV only
Explanation: Statement I is false. Being dimensionless means only that the coefficient has no physical unit; it does not make its numerical value universal. The coefficient depends on the interacting surface pair and their condition. Statement II is correct because
\[
f_{s,\max}=\mu_sN
\]
and
\[
f_k\approx\mu_kN
\]
both contain the normal reaction. Statement III is also correct because static friction is self-adjusting and can take any required value from zero up to its limit. If no horizontal force tends to produce relative motion, the required static friction can be zero. Statement IV is correct because elementary friction laws are empirical approximations rather than unlimited exact laws. Thus II, III and IV give the scientifically appropriate set.
617. A single particle is tested in four uniform-circular-motion trials. The measured values are:
| Trial | Speed \(v\) | Radius \(r\) | Inward net force \(F_c\) |
|---|
| P | \(4\,\text{m s}^{-1}\) | \(2\,\text{m}\) | \(8\,\text{N}\) |
| Q | \(6\,\text{m s}^{-1}\) | \(3\,\text{m}\) | \(12\,\text{N}\) |
| R | \(5\,\text{m s}^{-1}\) | \(5\,\text{m}\) | \(5\,\text{N}\) |
| S | \(3\,\text{m s}^{-1}\) | \(1.5\,\text{m}\) | \(9\,\text{N}\) |
If the mass of the particle is unchanged in all trials, which measurement is inconsistent with the other three?
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: For uniform circular motion,
\[
F_c=\frac{mv^2}{r}.
\]
Hence each trial implies
\[
m=\frac{F_cr}{v^2}.
\]
For P,
\[
m_P=\frac{8(2)}{4^2}=1\,\text{kg}.
\]
For Q,
\[
m_Q=\frac{12(3)}{6^2}=1\,\text{kg}.
\]
For R,
\[
m_R=\frac{5(5)}{5^2}=1\,\text{kg}.
\]
For S,
\[
m_S=\frac{9(1.5)}{3^2}=1.5\,\text{kg}.
\]
Thus P, Q and R are mutually consistent with the same \(1\,\text{kg}\) particle, whereas S would require a different mass. Therefore Trial S is the inconsistent measurement.
618. A \(2\,\text{kg}\) particle moves uniformly in a circle of radius \(2\,\text{m}\) with speed \(3\,\text{m s}^{-1}\). After one complete revolution, which statement is correct?
ⓐ. Net impulse over one revolution \(=0\); average net force \(=0\); instantaneous net force is zero at the starting point.
ⓑ. Net impulse over one revolution \(=18\,\text{N s}\); average net force \(=9\,\text{N}\); instantaneous magnitude \(=9\,\text{N}\).
ⓒ. Net impulse over one revolution \(=0\); average net force \(=0\); instantaneous net-force magnitude \(=9\,\text{N}\).
ⓓ. Net impulse over one revolution is non-zero; average net-force magnitude \(=9\,\text{N}\) because the instantaneous magnitude is constant.
Correct Answer: Net impulse over one revolution \(=0\); average net force \(=0\); instantaneous net-force magnitude \(=9\,\text{N}\).
Explanation: The instantaneous inward force is
\[
F_c=\frac{mv^2}{r}.
\]
Therefore,
\[
F_c=\frac{2(3^2)}2=9\,\text{N}.
\]
Its direction changes continuously as the particle moves around the circle. After one complete revolution, the particle returns with exactly the same velocity vector with which it started. Hence
\[
\Delta\vec p=0.
\]
The net impulse over the complete revolution is therefore
\[
\vec J=\Delta\vec p=0.
\]
Consequently,
\[
\vec F_{\text{avg}}=\frac{\Delta\vec p}{\Delta t}=0.
\]
Thus a non-zero force acting continuously can have zero vector impulse over a full cycle because its direction varies.
619. A particle of fixed mass moves uniformly in a circle of radius \(R\) with period \(T\) under an inward resultant of constant magnitude \(F\). In a second experiment the inward resultant has the same magnitude \(F\), but the radius is \(4R\). What must be the new period and new linear speed?
ⓐ. \(T'=2T,\ v'=v/2\)
ⓑ. \(T'=4T,\ v'=v\)
ⓒ. \(T'=T/2,\ v'=8v\)
ⓓ. \(T'=2T,\ v'=2v\)
Correct Answer: \(T'=2T,\ v'=2v\)
Explanation: In terms of period,
\[
F=\frac{4\pi^2mR}{T^2}.
\]
The mass and inward-force magnitude are unchanged. Therefore
\[
\frac{R}{T^2}=\text{constant}.
\]
When
\[
R'=4R,
\]
we require
\[
\frac{4R}{T'^2}=\frac{R}{T^2}.
\]
Hence,
\[
T'^2=4T^2,
\]
so
\[
T'=2T.
\]
The linear speed is
\[
v=\frac{2\pi R}{T}.
\]
Thus,
\[
v'=\frac{2\pi(4R)}{2T}=2v.
\]
The larger radius is therefore accompanied by both a longer period and a larger linear speed while preserving the same inward force.
620. At a point \(P\) on a circular path, a particle has velocity directed due east while its instantaneous net force is directed due north. The motion is uniform. Which statement correctly describes the geometry and the motion one-quarter revolution later?
ⓐ. The centre lies south of \(P\); the motion is anticlockwise, and after one-quarter revolution the net force is eastward.
ⓑ. The centre lies east of \(P\); the motion is clockwise, and after one-quarter revolution the velocity is southward.
ⓒ. The centre lies north of \(P\); the motion is anticlockwise, and after one-quarter revolution the velocity is northward while the net force is westward.
ⓓ. The centre lies north of \(P\); after one-quarter revolution both velocity and force point westward.
Correct Answer: The centre lies north of \(P\); the motion is anticlockwise, and after one-quarter revolution the velocity is northward while the net force is westward.
Explanation: In uniform circular motion, the net force is always directed radially inward. Since the net force at \(P\) points north, the centre of the circle must lie north of \(P\). Thus \(P\) is the southernmost point of the circular path. At that point an eastward tangent corresponds to anticlockwise motion. After one-quarter revolution, the particle reaches the easternmost point of the circle. The tangent there in anticlockwise motion points northward. The centre is then to the west of the particle. Therefore the inward net force points westward. Velocity and force remain perpendicular at each instant in this uniform circular motion.
