301. A \(10\,\text{kg}\) block is on a horizontal rough surface. A horizontal force of \(35\,\text{N}\) is applied, and the limiting friction is \(40\,\text{N}\). The block
ⓐ. moves with kinetic friction \(40\,\text{N}\)
ⓑ. remains at rest with static friction \(40\,\text{N}\)
ⓒ. moves with acceleration \(3.5\,\text{m s}^{-2}\)
ⓓ. remains at rest with static friction \(35\,\text{N}\)
Correct Answer: remains at rest with static friction \(35\,\text{N}\)
Explanation: \( \textbf{Given:} \) Applied force \(F=35\,\text{N}\) and limiting friction \(f_{s,\max}=40\,\text{N}\).
The block will remain at rest if static friction can balance the applied force.
The required static friction is \(35\,\text{N}\).
Since:
\[35\,\text{N}\lt40\,\text{N}\]
static friction can adjust to \(35\,\text{N}\).
It acts opposite to the applied force.
The value \(40\,\text{N}\) is only the maximum possible static friction, not the actual friction here.
\( \textbf{Final answer:} \) The block remains at rest with static friction \(35\,\text{N}\).
302. A \(5\,\text{kg}\) block on a rough horizontal surface has \(\mu_s=0.40\) and \(\mu_k=0.25\). A horizontal pull of \(30\,\text{N}\) is applied. Taking \(g=10\,\text{m s}^{-2}\), the block’s initial acceleration after it starts sliding is
ⓐ. \(2.0\,\text{m s}^{-2}\)
ⓑ. \(6.0\,\text{m s}^{-2}\)
ⓒ. \(3.5\,\text{m s}^{-2}\)
ⓓ. \(1.0\,\text{m s}^{-2}\)
Correct Answer: \(3.5\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(m=5\,\text{kg}\), \(\mu_s=0.40\), \(\mu_k=0.25\), \(F=30\,\text{N}\), and \(g=10\,\text{m s}^{-2}\).
\( \textbf{Normal reaction:} \)
\[N=mg=(5)(10)=50\,\text{N}\]
\( \textbf{Maximum static friction:} \)
\[f_{s,\max}=\mu_sN=(0.40)(50)=20\,\text{N}\]
Since \(30\,\text{N}\gt20\,\text{N}\), the block starts sliding.
After sliding begins, kinetic friction is used:
\[f_k=\mu_kN=(0.25)(50)=12.5\,\text{N}\]
\( \textbf{Net force during sliding:} \)
\[\sum F=30-12.5=17.5\,\text{N}\]
\( \textbf{Acceleration:} \)
\[a=\frac{17.5}{5}=3.5\,\text{m s}^{-2}\]
\( \textbf{Final answer:} \) The acceleration is \(3.5\,\text{m s}^{-2}\).
303. A horizontal pull is applied to a block on a rough floor. The block is just about to move. At this instant, the friction force is
ⓐ. maximum kinetic friction
ⓑ. zero friction
ⓒ. limiting static friction
ⓓ. rolling friction only
Correct Answer: limiting static friction
Explanation: When the block is just about to move, sliding has not yet started. Therefore the friction is still static friction. But it has reached its maximum possible value, called limiting static friction. This value is written as \(f_{s,\max}=\mu_sN\). Kinetic friction acts only after relative sliding begins. The phrase “just about to move” is the condition that selects limiting friction rather than ordinary adjustable static friction.
304. A block is pulled on a rough horizontal surface by a force \(F\) making angle \(\theta\) above the horizontal. The normal reaction is best written as
ⓐ. \(N=mg+F\sin\theta\)
ⓑ. \(N=mg-F\cos\theta\)
ⓒ. \(N=mg-F\sin\theta\)
ⓓ. \(N=F\sin\theta-mg\)
Correct Answer: \(N=mg-F\sin\theta\)
Explanation: The applied force has an upward vertical component \(F\sin\theta\). The weight \(mg\) acts downward, and the normal reaction \(N\) acts upward. If there is no vertical acceleration, the vertical force balance is used. Taking upward as positive:
\[N+F\sin\theta-mg=0\]
Rearranging gives:
\[N=mg-F\sin\theta\]
The pull partly lifts the block, so the surface has to push less strongly. Using \(mg+F\sin\theta\) would describe a downward component pressing the block harder into the surface, not an upward pull.
305. A block is pushed on a rough horizontal surface by a force \(F\) making angle \(\theta\) below the horizontal. The maximum static friction becomes
ⓐ. \(\mu_s(mg-F\sin\theta)\)
ⓑ. \(\mu_s(mg+F\sin\theta)\)
ⓒ. \(\mu_sF\cos\theta\)
ⓓ. \(\mu_smg-F\cos\theta\)
Correct Answer: \(\mu_s(mg+F\sin\theta)\)
Explanation: A push inclined below the horizontal has a downward vertical component \(F\sin\theta\). This increases the normal reaction because the surface must support both the weight and the extra downward component. With no vertical acceleration:
\[N=mg+F\sin\theta\]
The maximum static friction is:
\[f_{s,\max}=\mu_sN\]
Substituting \(N\):
\[f_{s,\max}=\mu_s(mg+F\sin\theta)\]
The horizontal component \(F\cos\theta\) tends to move the block, but the vertical component controls how large the normal reaction and friction limit become.
306. A \(20\,\text{N}\) force pulls a \(5\,\text{kg}\) block at \(30^\circ\) above the horizontal on a rough horizontal floor. Taking \(g=10\,\text{m s}^{-2}\), the normal reaction is
ⓐ. \(50\,\text{N}\)
ⓑ. \(60\,\text{N}\)
ⓒ. \(10\,\text{N}\)
ⓓ. \(40\,\text{N}\)
Correct Answer: \(40\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(F=20\,\text{N}\), \(\theta=30^\circ\), \(m=5\,\text{kg}\), and \(g=10\,\text{m s}^{-2}\).
\( \textbf{Weight:} \)
\[mg=(5)(10)=50\,\text{N}\]
The upward vertical component of pull is:
\[F\sin30^\circ=(20)\left(\frac{1}{2}\right)=10\,\text{N}\]
With no vertical acceleration:
\[N+F\sin30^\circ-mg=0\]
So:
\[N=mg-F\sin30^\circ\]
\[N=50-10=40\,\text{N}\]
The pull reduces the normal reaction because part of the force acts upward.
\( \textbf{Final answer:} \) The normal reaction is \(40\,\text{N}\).
307. A \(20\,\text{N}\) force pushes a \(5\,\text{kg}\) block at \(30^\circ\) below the horizontal on a rough horizontal floor. Taking \(g=10\,\text{m s}^{-2}\), the normal reaction is
ⓐ. \(50\,\text{N}\)
ⓑ. \(60\,\text{N}\)
ⓒ. \(40\,\text{N}\)
ⓓ. \(10\,\text{N}\)
Correct Answer: \(60\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(F=20\,\text{N}\), \(\theta=30^\circ\), \(m=5\,\text{kg}\), and \(g=10\,\text{m s}^{-2}\).
\( \textbf{Weight:} \)
\[mg=(5)(10)=50\,\text{N}\]
The downward vertical component of the push is:
\[F\sin30^\circ=(20)\left(\frac{1}{2}\right)=10\,\text{N}\]
The surface must support the weight and this extra downward component.
\( \textbf{Vertical balance:} \)
\[N-mg-F\sin30^\circ=0\]
So:
\[N=mg+F\sin30^\circ\]
\[N=50+10=60\,\text{N}\]
The downward push increases the normal reaction and can therefore increase the frictional limit.
\( \textbf{Final answer:} \) The normal reaction is \(60\,\text{N}\).
308. A table compares two ways of applying the same oblique force \(F\) to a block on a rough horizontal floor.
| Case | Force direction | Effect on normal reaction |
|---|
| P | Pulling above the horizontal | Decreases \(N\) |
| Q | Pushing below the horizontal | Increases \(N\) |
| R | Pulling above the horizontal | Always makes \(N=mg\) |
| S | Pushing below the horizontal | Always makes \(N=0\) |
The fully suitable rows are
ⓐ. Q and R only
ⓑ. P and Q only
ⓒ. R and S only
ⓓ. P and S only
Correct Answer: P and Q only
Explanation: When a force pulls above the horizontal, it has an upward vertical component. This reduces the normal reaction because the surface has to support less of the downward load. When a force pushes below the horizontal, it has a downward vertical component. This increases the normal reaction because the surface must push harder upward. Row R is unsuitable because \(N=mg\) only in special cases with no other vertical force. Row S is unsuitable because pushing downward cannot make the normal reaction zero; it generally increases \(N\).
309. A block on a rough horizontal surface is pulled by a force \(F\) at angle \(\theta\) above the horizontal. If it is just about to move, the horizontal balance condition is
ⓐ. \(F\cos\theta=\mu_s(mg-F\sin\theta)\)
ⓑ. \(F\sin\theta=\mu_s(mg-F\cos\theta)\)
ⓒ. \(F=\mu_smg\) always
ⓓ. \(F\cos\theta=mg+\mu_sF\)
Correct Answer: \(F\cos\theta=\mu_s(mg-F\sin\theta)\)
Explanation: At impending motion, static friction reaches its limiting value. The horizontal component trying to move the block is \(F\cos\theta\). Since the force pulls upward, the normal reaction is \(N=mg-F\sin\theta\). The limiting friction is:
\[f_{s,\max}=\mu_sN=\mu_s(mg-F\sin\theta)\]
For the block to be just about to move:
\[F\cos\theta=f_{s,\max}\]
Therefore:
\[F\cos\theta=\mu_s(mg-F\sin\theta)\]
This equation keeps the horizontal driving component and the vertical effect on normal reaction separate.
310. A \(4\,\text{kg}\) block is pulled along a rough horizontal surface by a horizontal force of \(25\,\text{N}\). If \(\mu_k=0.50\) and \(g=10\,\text{m s}^{-2}\), its acceleration is
ⓐ. \(1.25\,\text{m s}^{-2}\)
ⓑ. \(6.25\,\text{m s}^{-2}\)
ⓒ. \(5.00\,\text{m s}^{-2}\)
ⓓ. \(0.50\,\text{m s}^{-2}\)
Correct Answer: \(1.25\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(m=4\,\text{kg}\), \(F=25\,\text{N}\), \(\mu_k=0.50\), and \(g=10\,\text{m s}^{-2}\).
\( \textbf{Normal reaction:} \)
\[N=mg=(4)(10)=40\,\text{N}\]
\( \textbf{Kinetic friction:} \)
\[f_k=\mu_kN=(0.50)(40)=20\,\text{N}\]
\( \textbf{Net horizontal force:} \)
\[\sum F=25-20=5\,\text{N}\]
\( \textbf{Acceleration:} \)
\[a=\frac{\sum F}{m}\]
\[a=\frac{5}{4}=1.25\,\text{m s}^{-2}\]
The applied force is only slightly larger than kinetic friction, so the acceleration is much smaller than \(\frac{25}{4}\,\text{m s}^{-2}\).
\( \textbf{Final answer:} \) The acceleration is \(1.25\,\text{m s}^{-2}\).
311. On an inclined plane of angle \(\theta\), the component of weight parallel to the plane is
ⓐ. \(mg\sin\theta\)
ⓑ. \(mg\cos\theta\)
ⓒ. \(mg\tan\theta\)
ⓓ. \(\frac{mg}{\sin\theta}\)
Correct Answer: \(mg\sin\theta\)
Explanation: The weight \(mg\) acts vertically downward. When axes are chosen parallel and perpendicular to an inclined plane, the weight is resolved into two components. The component along the plane is \(mg\sin\theta\), directed down the plane. The component perpendicular to the plane is \(mg\cos\theta\), directed into the plane. These components are based on the plane’s angle \(\theta\) with the horizontal. Mixing sine and cosine gives wrong friction and acceleration equations on an incline.
312. A block rests on a smooth inclined plane of angle \(\theta\). The normal reaction is
ⓐ. \(mg\sin\theta\)
ⓑ. \(mg\tan\theta\)
ⓒ. \(mg\cos\theta\)
ⓓ. \(mg\)
Correct Answer: \(mg\cos\theta\)
Explanation: On an inclined plane, the normal reaction is perpendicular to the plane. The component of weight perpendicular to the plane is \(mg\cos\theta\). If the block has no acceleration perpendicular to the plane, the normal reaction balances this perpendicular component. Therefore \(N=mg\cos\theta\). The component \(mg\sin\theta\) acts parallel to the plane and tends to make the block slide down. The full weight \(mg\) is vertical, not perpendicular to the inclined surface.
313. A block is placed on a rough inclined plane and tends to slide down. The static friction on the block acts
ⓐ. down the plane
ⓑ. perpendicular into the plane
ⓒ. up the plane
ⓓ. vertically downward
Correct Answer: up the plane
Explanation: Static friction opposes the tendency of relative motion between surfaces. On the incline, gravity has a component \(mg\sin\theta\) down the plane. If this makes the block tend to slide down, friction must act up the plane. The normal reaction acts perpendicular to the plane, not along it. Weight acts vertically downward and is not the friction force. The direction of friction is decided by the likely slipping direction, not by simply saying friction always points left or right.
314. A \(10\,\text{kg}\) block is on a rough inclined plane of angle \(30^\circ\). It is just at rest and about to slide down. Taking \(g=10\,\text{m s}^{-2}\), the limiting friction is
ⓐ. \(50\,\text{N}\) up the plane
ⓑ. \(50\,\text{N}\) down the plane
ⓒ. \(100\,\text{N}\) up the plane
ⓓ. \(86.6\,\text{N}\) down the plane
Correct Answer: \(50\,\text{N}\) up the plane
Explanation: \( \textbf{Given:} \) \(m=10\,\text{kg}\), \(\theta=30^\circ\), and \(g=10\,\text{m s}^{-2}\).
\( \textbf{Weight:} \)
\[mg=(10)(10)=100\,\text{N}\]
The component down the plane is:
\[mg\sin30^\circ=100\left(\frac{1}{2}\right)=50\,\text{N}\]
The block is just at rest and about to slide down, so limiting friction acts up the plane.
For equilibrium at the limiting condition:
\[f_{s,\max}=mg\sin30^\circ\]
\[f_{s,\max}=50\,\text{N}\]
The direction is up the plane because friction opposes the downward tendency of motion.
\( \textbf{Final answer:} \) The limiting friction is \(50\,\text{N}\) up the plane.
315. For a block just about to slide down a rough inclined plane of angle \(\theta\), the coefficient of static friction is related to the angle by
ⓐ. \(\mu_s=\sin\theta\)
ⓑ. \(\mu_s=\cos\theta\)
ⓒ. \(\mu_s=\tan\theta\)
ⓓ. \(\mu_s=\frac{1}{\tan\theta}\) always
Correct Answer: \(\mu_s=\tan\theta\)
Explanation: At the limiting condition, the block is just about to slide down the plane. The down-plane component of weight is \(mg\sin\theta\). The normal reaction is \(N=mg\cos\theta\). The limiting friction is \(f_{s,\max}=\mu_sN=\mu_smg\cos\theta\), acting up the plane. For equilibrium at the limiting condition:
\[mg\sin\theta=\mu_smg\cos\theta\]
Cancel \(mg\):
\[\sin\theta=\mu_s\cos\theta\]
So:
\[\mu_s=\tan\theta\]
This relation applies at the angle of repose, not at every arbitrary incline angle.
316. The angle of repose is the angle of an inclined plane at which a body
ⓐ. moves upward with maximum speed
ⓑ. has zero weight
ⓒ. is just about to slide down
ⓓ. experiences no normal reaction on a plane
Correct Answer: is just about to slide down
Explanation: The angle of repose is defined for a body on a rough inclined plane. It is the angle at which the body is just about to start sliding down under its own weight. At this condition, static friction is limiting. The relation \(\mu_s=\tan\theta_r\) holds, where \(\theta_r\) is the angle of repose. The body does not lose weight at this angle. The normal reaction is generally non-zero and equals \(mg\cos\theta_r\).
317. If the coefficient of static friction between a block and an inclined plane is \(\mu_s=\frac{1}{\sqrt{3}}\), the angle of repose is
ⓐ. \(45^\circ\)
ⓑ. \(30^\circ\)
ⓒ. \(60^\circ\)
ⓓ. \(90^\circ\)
Correct Answer: \(30^\circ\)
Explanation: \( \textbf{Given:} \) \(\mu_s=\frac{1}{\sqrt{3}}\).
For angle of repose:
\[\mu_s=\tan\theta_r\]
\( \textbf{Substitution:} \)
\[\tan\theta_r=\frac{1}{\sqrt{3}}\]
The standard angle with \(\tan\theta=\frac{1}{\sqrt{3}}\) is:
\[\theta_r=30^\circ\]
This is the angle at which the block is just about to slide down the rough plane.
The relation uses static friction at the limiting condition.
\( \textbf{Final answer:} \) The angle of repose is \(30^\circ\).
318. A graph description is given below.
For several pairs of dry surfaces, limiting friction \(f_{s,\max}\) is plotted on the vertical axis against normal reaction \(N\) on the horizontal axis. The graph is a straight line through the origin.
The slope of this graph represents
ⓐ. \(mg\)
ⓑ. \(\frac{1}{\mu_s}\)
ⓒ. \(\mu_s\)
ⓓ. acceleration due to gravity
Correct Answer: \(\mu_s\)
Explanation: Limiting static friction is related to normal reaction by:
\[f_{s,\max}=\mu_sN\]
If \(f_{s,\max}\) is on the vertical axis and \(N\) is on the horizontal axis, the graph has the form \(y=mx\). The slope is therefore \(\mu_s\). A straight line through the origin shows direct proportionality between limiting friction and normal reaction for the same pair of surfaces. The slope is dimensionless because both axes represent forces. The graph does not directly give \(mg\) unless the normal reaction is separately connected to weight in a particular setup.
319. A block slides down a rough inclined plane of angle \(\theta\) with coefficient of kinetic friction \(\mu_k\). The acceleration down the plane is
ⓐ. \(g(\cos\theta-\mu_k\sin\theta)\)
ⓑ. \(g(\sin\theta-\mu_k\cos\theta)\)
ⓒ. \(g(\sin\theta+\mu_k\cos\theta)\)
ⓓ. \(\mu_kg\sin\theta\)
Correct Answer: \(g(\sin\theta-\mu_k\cos\theta)\)
Explanation: The component of weight down the plane is \(mg\sin\theta\). Since the block slides down, kinetic friction acts up the plane. The normal reaction is \(N=mg\cos\theta\). Therefore kinetic friction is:
\[f_k=\mu_kN=\mu_kmg\cos\theta\]
Taking down the plane as positive, the net force is:
\[\sum F=mg\sin\theta-\mu_kmg\cos\theta\]
Using \(\sum F=ma\):
\[ma=mg\sin\theta-\mu_kmg\cos\theta\]
Cancel \(m\):
\[a=g(\sin\theta-\mu_k\cos\theta)\]
The friction term is subtracted because it opposes the downward sliding.
320. A \(2\,\text{kg}\) block slides down a rough plane inclined at \(30^\circ\). If \(\mu_k=\frac{1}{2\sqrt{3}}\) and \(g=10\,\text{m s}^{-2}\), its acceleration down the plane is
ⓐ. \(5.0\,\text{m s}^{-2}\)
ⓑ. \(7.5\,\text{m s}^{-2}\)
ⓒ. \(2.5\,\text{m s}^{-2}\)
ⓓ. \(10\,\text{m s}^{-2}\)
Correct Answer: \(2.5\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(\theta=30^\circ\), \(\mu_k=\frac{1}{2\sqrt{3}}\), and \(g=10\,\text{m s}^{-2}\).
For a block sliding down a rough incline:
\[a=g(\sin\theta-\mu_k\cos\theta)\]
Use:
\[\sin30^\circ=\frac{1}{2}\]
\[\cos30^\circ=\frac{\sqrt{3}}{2}\]
\( \textbf{Friction term:} \)
\[\mu_k\cos30^\circ=\left(\frac{1}{2\sqrt{3}}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4}\]
\( \textbf{Substitution:} \)
\[a=10\left(\frac{1}{2}-\frac{1}{4}\right)\]
\[a=10\left(\frac{1}{4}\right)\]
\[a=2.5\,\text{m s}^{-2}\]
The mass cancels because both gravity and frictional terms are proportional to \(m\).
\( \textbf{Final answer:} \) The acceleration is \(2.5\,\text{m s}^{-2}\) down the plane.
