Question 1

Three blocks,light connecting ropes,and a light frictionless pulley comprise a system,as shown in the figure.An external force of magnitude P is exerted downward on block A,causing block A to accelerate downward at a constant 2.5 m/s².The rope connecting block B and block C exerts a 60-N force. (a)What is the magnitude of the force P? (b)What is the mass of block C?

Accepted Answer

The answer of Three blocks,light connecting ropes,and a light frictionless...

Question 2

Jason takes off across level water on his jet-powered skis.The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible).Water exerts a force of 200 N on the skis.The coefficient of kinetic friction between skis and water is 0.10.Unfortunately,the skis run out of fuel after only 90 s.What is Jason's top speed?

Accepted Answer

The answer of Jason takes off across level water on...

Question 3

A flatbed truck is carrying a 20.0-kg crate along a level road.The coefficient of static friction between the crate and the bed is 0.400,and the coefficient of kinetic friction is 0.300.What is the maximum acceleration that the truck can have if the crate is to stay in place relative to the truck?

Accepted Answer

To keep the crate in place relative to the truck, the friction force must be equal and opposite to the force of the truck's acceleration. The maximum static friction force is given by $F_s = \mu_s N$, where $N$ is the normal force (equal to the weight of the crate) and $\mu_s$ is the coefficient of static friction. Therefore, the maximum acceleration the truck can have before the crate starts to slide is $a_{max}=\frac{\mu_s N}{m} = \frac{\mu_s g m}{m} = \mu_s g$. Plugging in the given values, we get $a_{max} = 0.400 \cdot 9.81 \approx 3.92$ m/s$^2$. Therefore, the best choice is A.

Question 4

A bulldozer attempts to drag a log weighing 500 N along the rough horizontal ground.The cable attached to the log makes an angle of 30° above the ground.The coefficient of static friction between the log and the ground is 0.50,and the coefficient of kinetic friction is 0.35.What minimum force should the cable exert on the log tension in order for the log to begin to slide?

Accepted Answer

The answer of A bulldozer attempts to drag a log...

Question 5

In the figure,block A has a mass of 3.00 kg.It rests on a smooth horizontal table and is connected by a very light horizontal string over an ideal pulley to block B,which has a mass of 2.00 kg.When block B is gently released from rest,how long does it take block B to travel 80.0 cm?

Accepted Answer

The answer of In the figure,block A has a mass...

Question 6

In a shuffleboard game,the puck slides a total of 12 m on a horizontal surface before coming to rest.If the coefficient of kinetic friction between the puck and board is 0.10,what was the initial speed of the puck?

Accepted Answer

The initial speed of the puck can be found using the work-energy principle. The work done by friction is equal to the change in kinetic energy. The work done by friction is $W = f_k \cdot d = \mu_k \cdot m \cdot g \cdot d$, where $f_k$ is the force of kinetic friction, $\mu_k$ is the coefficient of kinetic friction, $m$ is the mass of the puck, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $d$ is the distance (12 m). The change in kinetic energy is $\Delta KE = \frac{1}{2} m v^2 - 0$, since the puck comes to rest. Setting the work done by friction equal to the change in kinetic energy and solving for $v$ (initial speed), we don't actually need the mass of the puck as it cancels out. Using the given coefficient of kinetic friction ($\mu_k = 0.10$) and the distance (12 m), we can solve for the initial speed $v$. The equation simplifies to $v = \sqrt{2 \cdot \mu_k \cdot g \cdot d}$. Plugging in the values, $v = \sqrt{2 \cdot 0.10 \cdot 9.8 \cdot 12}$, which calculates to approximately $4.8 \, \text{m/s}$.

Question 7

During a hockey game,a puck is given an initial speed of 10 m/s.It slides 50 m on the horizontal ice before it stops due to friction.What is the coefficient of kinetic friction between the puck and the ice?

Accepted Answer

The answer of During a hockey game,a puck is given...

Question 8

You push downward on a trunk at an angle 25° below the horizontal with a force of $750 \mathrm {~N}$ If the trunk is on a flat surface and the coefficient of static friction between the surface and the trunk is 0.61,what is the most massive trunk you will be able to move?

Accepted Answer

The maximum force of static friction is:$$f_s=\mu_sn$$$$f_s=0.61n$$The normal force is:$$n=mg\cos25^\circ$$The force applied is:$$F=750	ext{ N}$$The angle between the force and the normal force is:$$	heta=90^\circ-25^\circ=65^\circ$$The maximum force of static friction is:$$f_s=\mu_sn=0.61mg\cos25^\circ$$The maximum force that can be applied without moving the trunk is:$$F=f_s=0.61mg\cos25^\circ$$The mass of the trunk is:$$m=\frac{F}{0.61g\cos25^\circ}=\frac{750	ext{ N}}{0.61(9.81	ext{ m/s}^2)\cos25^\circ}=93	ext{ kg}$$

Question 9

A 50-kg box is being pushed along a horizontal surface.The coefficient of static friction between the box and the ground is 0.65,and the coefficient of kinetic friction is 0.35.What horizontal force must be exerted on the box for it to accelerate at 1.2 m/s²?

Accepted Answer

The answer of A 50-kg box is being pushed along...

Question 10

A locomotive is pulling three train cars along a level track exerting a force of 100,000 N.The car next to the locomotive has a mass of 80,000 kg,the next one,50,000 kg,and the last one,70,000 kg.You can neglect the friction on the cars being pulled.
(a)What is the magnitude of the force that the 80,000-kg car exerts on the 50,000-kg car?
(b)What is the magnitude of the force that the 50,000-kg car exerts on the 70,000-kg car?

Accepted Answer

The answer of A locomotive is pulling three train cars...

Question 11

A horizontal 52-N force must be exerted for a $50 - \mathrm { kg }$ box to slide across a flat surface at a constant velocity of 3.5 m/s.What is the coefficient of kinetic friction between the box and the floor?

Accepted Answer

The coefficient of kinetic friction ($\mu_k$) is calculated using the formula $\mu_k = \frac{F}{N}$, where $F$ is the force of friction (52 N in this case) and $N$ is the normal force. For an object on a flat surface, the normal force equals the weight of the object, which is $mg = 50 \times 9.8 = 490 \, \text{N}$. Thus, $\mu_k = \frac{52}{490} \approx 0.106$, which rounds to 0.11.

Question 12

An object slides on a level floor.It slows and comes to a stop with a constant acceleration of magnitude 2.4 m/s².What is the coefficient of kinetic friction between the object and the floor?

Accepted Answer

The answer of An object slides on a level floor.It...

Question 13

The coefficients of static and kinetic friction between a 3.0-kg box and a horizontal desktop are 0.40 and 0.30,respectively.What is the force of friction that the desktop exerts on the box when a 15-N horizontal push is exerted on the box?

Accepted Answer

The force of friction can be calculated as follows:
If the box is at rest, the force of friction will be equal and opposite to the applied force, 15 N. Therefore, frictional force = 15 N.
However, if the box is moving, we need to use the coefficients of friction to calculate the force of friction. The maximum frictional force will be equal to the product of the coefficient of static friction and the normal force, i.e., F_friction,max = μ_static x (m x g) = 0.40 x (3.0 kg x 9.8 m/s^2) = 11.76 N. Since the applied force of 15 N is greater than the maximum static frictional force, the box will start moving and the force of friction will be equal to the product of the coefficient of kinetic friction and the normal force, i.e., F_friction = μ_kinetic x (m x g) = 0.30 x (3.0 kg x 9.8 m/s^2) = 8.82 N. Therefore, the force of friction that the desktop exerts on the box is 8.82 N, which is closest to option B) 8.8 N.

Question 14

The figure shows a 100-kg block being released from rest from a height of 1.0 m.It then takes 0.53 s for it to reach the floor.What is the mass m of the block on the left? There is no friction or mass in the pulley,and the connecting rope is very light.

Accepted Answer

The answer of The figure shows a 100-kg block being...

Question 15

A 55-kg box rests on a horizontal surface.The coefficient of static friction between the box and the surface is 0.30,and the coefficient of kinetic friction is 0.20.What horizontal force must be applied to the box to cause it to start sliding along the surface?

Accepted Answer

The answer of A 55-kg box rests on a horizontal...

Question 16

A baseball player is running to second base at 5.03 m/s.When he is 4.80 m from the plate he goes into a slide.The coefficient of kinetic friction between the player and the ground is 0.180,and the coefficient of static friction is 3.14.What is his speed when he reaches the plate?

Accepted Answer

The answer of A baseball player is running to second...

Question 17

A 50-kg box is resting on a horizontal floor.A force of 250 N directed at an angle of 30.0° below the horizontal is applied to the box.The coefficient of static friction between the box and the surface is 0.40,and the coefficient of kinetic friction is 0.30.What is the force of friction exerted by the floor on the box?

Accepted Answer

First, we need to resolve the force into horizontal and vertical components:
F_horizontal = Fcos(30) = 250cos(30) = 216.5 N
F_vertical = Fsin(30) = 250sin(30) = 125 N

Next, we need to find the maximum force of static friction:
f_static = μ_static * N
N = mg = 50 * 9.8 = 490 N
f_static = 0.40 * 490 = 196 N

Since the applied force is greater than the maximum force of static friction, the box will start moving and the force of friction will be the force of kinetic friction:
f_kinetic = μ_kinetic * N
f_kinetic = 0.30 * 490 = 147 N

Therefore, the force of friction exerted by the floor on the box is 147 N, which is closest to choice C (220 N).

Question 18

A 55-kg box rests on a horizontal surface.The coefficient of static friction between the box and the surface is 0.30.A horizontal 140-N force is exerted on the box.What is the friction force exerted by the surface on the box?

Accepted Answer

The friction force exerted by the surface on the box equals the applied force (140 N) because it is less than the maximum static friction force, which can be calculated as the product of the coefficient of static friction (0.30) and the normal force (the weight of the box, which is mass times gravity, 55 kg * 9.8 m/s^2 = 539 N). The maximum static friction force is 0.30 * 539 N = 161.7 N, which is greater than the applied force, meaning the box does not move, and the frictional force matches the applied force to prevent motion.

Question 19

A policeman investigating an accident measures the skid marks left by a car on the horizontal road.He determines that the distance between the point that the driver slammed on the brakes (thereby locking the wheels)and the point where the car came to a stop was 28.0 m.From a reference manual he determines that the coefficient of kinetic friction between the tires and the road under the prevailing conditions was 0.300.How fast was the car going when the driver applied the brakes?

Accepted Answer

The speed of the car when the brakes were applied can be calculated using the formula derived from the work-energy principle, $v = \sqrt{2ad}$, where $v$ is the velocity, $a$ is the acceleration (deceleration in this case), and $d$ is the distance. The deceleration $a$ can be found using the formula $a = \mu g$, where $\mu$ is the coefficient of kinetic friction (0.300) and $g$ is the acceleration due to gravity (approximately 9.8 m/s$^2$). Substituting the given values, we get $a = 0.300 \times 9.8 = 2.94$ m/s$^2$, and then $v = \sqrt{2 \times 2.94 \times 28.0} \approx 12.8$ m/s.

Question 20

A driver in a $1000 - \mathrm { kg }$ car traveling at $24 \mathrm {~m} / \mathrm { s }$ slams on the brakes and skids to a stop.If the coefficient of friction between the tires and the level road is 0.80,how long will the skid marks be?

Accepted Answer

The length of the skid marks can be calculated using the formula derived from the work-energy principle, where the work done by friction is equal to the change in kinetic energy. The formula is $d = \frac{v^2}{2\mu g}$, where $d$ is the distance (length of skid marks), $v$ is the initial velocity, $\mu$ is the coefficient of friction, and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Plugging in the given values: $d = \frac{(24)^2}{2 \times 0.80 \times 9.8} \approx 37 \, \text{m}$.

Quiz 4: Applying Newtons Laws