Question 1

A flatbed truck is carrying a 20-kg crate up a sloping road.The coefficient of static friction between the crate and the bed is 0.40,and the coefficient of kinetic friction is 0.30.What is the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place?
A)0.38°
B)13°
C)17°
D)22°
E)68°

Accepted Answer

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Question 2

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?
A)0.090
B)0.10
C)0.11
D)0.12

Accepted Answer

The work done by friction is equal to the initial kinetic energy of the puck. Using the work-energy principle and the formula for frictional force, we can solve for the coefficient of kinetic friction: $ \mu_k = \frac{v^2}{2gd} $. Substituting the given values, $ \mu_k = \frac{10^2}{2 \times 9.8 \times 50} = 0.10 $.

Question 3

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 on the box?
A)32 N
B)200 N
C)220 N
D)250 N
E)620 N

Accepted Answer

First, we need to resolve the applied force into horizontal and vertical components. The vertical component does not affect the motion of the box since it is perpendicular to the surface.

The horizontal component is 250N*cos(30°) = 216.5 N.

Since the box is not moving, the force of static friction must be equal in magnitude and opposite in direction to the horizontal component of the applied force. Therefore, the force of static friction is also 216.5 N.

The question asks for the force of friction, which would be the force of kinetic friction if the box were moving. Since the box is not moving, we use the coefficient of static friction and the force of static friction to find the force of friction:

Force of friction = coefficient of static friction * normal force 
Normal force = weight of box = 50 kg * 9.81 m/s^2 = 490.5 N 
Force of friction = 0.40 * 490.5 N = 196.2 N

Therefore, the force of friction on the box is 196.2 N, which is closest to choice C.

Question 4

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).The skis produce a forward thrust of 200 N and have a coefficient of kinetic friction with water of 0.10.Unfortunately,the skis run out of fuel after only 41 s.How far from his starting point has Jason traveled when he finally coasts to a stop?
A)3,900 m
B)2,900 m
C)2,000 m
D)1,400 m

Accepted Answer

The net force while the skis are powered is 200 N - (0.10 * 75 kg * 9.8 m/s²) = 127.5 N. The acceleration is 127.5 N / 75 kg = 1.7 m/s². Using the formula d = v₀t + 0.5at², with v₀ = 0, t = 41 s, and a = 1.7 m/s², the distance traveled while powered is 0.5 * 1.7 m/s² * (41 s)² = 1,430.35 m. The final velocity when the fuel runs out is v = at = 1.7 m/s² * 41 s = 69.7 m/s. When coasting to a stop, the deceleration is due to friction: a = -0.10 * 9.8 m/s² = -0.98 m/s². Using v² = u² + 2as, where v = 0, u = 69.7 m/s, and a = -0.98 m/s², the stopping distance is s = (69.7 m/s)² / (2 * 0.98 m/s²) = 2,478.57 m. The total distance is 1,430.35 m + 2,478.57 m = 3,908.92 m, approximately 3,900 m.

Question 5

The figure shows a block of mass m resting on a 20° slope.The block has coefficients of friction 
And 
With the surface of the slope.It is connected using a very light string over an ideal pulley to a hanging block of mass 2.0 kg.The string above the slope pulls parallel to the surface.What is the minimum mass m so the system will remain at rest when it is released from rest? 
A)2.1 kg
B)3.3 kg
C)1.3 kg
D)3.6 kg

Accepted Answer

The answer of The figure shows a block of mass...

Question 6

A 200-g hockey puck is launched at an initial speed of 16 m/s up a metal ramp that is inclined at a 30° angle.The coefficients of static and kinetic friction between the hockey puck and the metal ramp are µ_s = 0.40 and µ_k = 0.30,respectively.What vertical height does the puck reach above its starting point? A)8.6 cm B)17 cm C)13 cm D)4.2 cm

Accepted Answer

The correct answer is A because the puck will reach a vertical height of 8.6 cm above its starting point.

Question 7

A horizontal 52-N force is needed to slide a 
Box 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?
A)0.11
B)0.09
C)0.10
D)0.13

Accepted Answer

The answer of A horizontal 52-N force is needed to...

Question 8

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 applied to the box.What is the friction force on the box?
A)140 N
B)160 N
C)16.5 N
D)0.00 N
E)42 N

Accepted Answer

The friction force equals the applied force (140 N) because it's 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 (mass * gravity, 55 kg * 9.8 m/s^2 = 539 N). The maximum static friction force is 0.30 * 539 N = 161.7 N, and since 140 N is less than this, the friction force matches the applied force to prevent motion.

Question 9

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?
A)4.8 m/s
B)48.5 m/s
C)3.8 m/s
D)4.3 m/s

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, the final kinetic energy is 0). Setting the work done by friction equal to the change in kinetic energy and solving for $v$ (initial speed), we don't need the mass of the puck as it cancels out. Plugging in the values, we get $v = \sqrt{2 \cdot \mu_k \cdot g \cdot d}$. Substituting $\mu_k = 0.10$, $g = 9.8 \, \text{m/s}^2$, and $d = 12 \, \text{m}$, we find $v \approx 4.8 \, \text{m/s}$.

Question 10

A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N that is parallel to the slope.The coefficient of kinetic friction between the block and the slope is 0.200.What is the acceleration of the block? A)0.528 m/ s² B)0.158 m/ s² C)0.412 m/ s² D)0.983 m/ s² E)0.260 m/ s²

Accepted Answer

The answer of A 50.0-kg block is being pulled up...

Question 11

An ornament of mass 40.0 g is attached to a vertical ideal spring with a force constant (spring constant)of 20.0 N/m.The ornament is then lowered very slowly until the spring stops stretching.How much does the spring stretch?
A)0.00200 m
B)0.0196 m
C)0.0816 m
D)0.800 m
E)0.200 m

Accepted Answer

We can use Hooke's Law, which states that the force exerted by the spring is equal to the spring constant times the displacement of the spring from its equilibrium position: F = -kx (- sign indicates that the force is in the opposite direction of the displacement). At equilibrium, the force exerted by gravity on the mass is balanced by the force exerted by the spring, so we can write: Fg = -kx. Solving for x, we get: x = -Fg/k = -(mg)/k = - (0.040 kg)(9.81 m/s^2)/(20.0 N/m) = -0.0196 m. The negative sign is just indicating that the displacement is in the opposite direction of the gravitational force, so we take the absolute value to get: x = 0.0196 m.

Question 12

A driver in a 
Car traveling at 
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?
A)37 m
B)30 m
C)46 m
D)34 m

Accepted Answer

The answer of A driver in a 
Car traveling at...

Question 13

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).The skis produce a forward thrust of 200 N and have a coefficient of kinetic friction with water of 0.10.Unfortunately,the skis run out of fuel after only 90 s.What is Jason's top speed?
A)150 m/s
B)240 m/s
C)24 m/s
D)90 m/s

Accepted Answer

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

Question 14

You push downward on a trunk at an angle 25° below the horizontal with a force of 
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?
A)81 kg
B)93 kg
C)73 kg
D)112 kg

Accepted Answer

The answer of You push downward on a trunk at...

Question 15

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?
A)4.47 m/s
B)2.89 m/s
C)1.96 m/s
D)2.56 m/s
E)He stops before reaching the plate.

Accepted Answer

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

Question 16

A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal.If the coefficient of kinetic friction is 0.40,at what rate does the box accelerate down the slope? A)3.1 m/s² B)3.4 m/s² C)3.7 m/s² D)4.1 m/s²

Accepted Answer

The correct answer is A because the acceleration of the box down the slope is given by a = g * sin(theta) - mu * g * cos(theta), where g is the acceleration due to gravity, theta is the angle of the incline, and mu is the coefficient of kinetic friction. Plugging in the given values, we get a = 9.8 m/s^2 * sin(39°) - 0.40 * 9.8 m/s^2 * cos(39°) = 3.1 m/s^2.

Question 17

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?
A)10.7 m/s
B)12.8 m/s
C)21.4 m/s
D)32.9 m/s
E)45.7 m/s

Accepted Answer

The speed of the car when the brakes were applied can be calculated using the formula derived from the work-energy principle, $v = \sqrt{2\mu g d}$, where $v$ is the initial velocity, $\mu$ is the coefficient of kinetic friction (0.300), $g$ is the acceleration due to gravity (approximately 9.8 m/s$^2$), and $d$ is the distance (28.0 m). Plugging in the values, $v = \sqrt{2 * 0.300 * 9.8 * 28.0}$ gives a result close to 12.8 m/s.

Question 18

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? A)60 N B)120 N C)170 N D)230 N E)490 N

Accepted Answer

To find the force needed to accelerate the box, first calculate the force of kinetic friction: F_friction = μ_k * m * g = 0.35 * 50 kg * 9.8 m/s² = 171.5 N. Then, calculate the net force required for the acceleration: F_net = m * a = 50 kg * 1.2 m/s² = 60 N. The total force needed is the sum of the frictional force and the net force: F_total = F_friction + F_net = 171.5 N + 60 N = 231.5 N, which rounds to 230 N.

Question 19

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? A)3.92 m/s² B)7.84 m/s² C)8.00 m/s² D)78.5 m/s² E)196 m/s²

Accepted Answer

The maximum acceleration that the truck can have if the crate is to stay in place is equal to the maximum static friction force between the crate and the bed, which is μsN, where μs is the coefficient of static friction and N is the normal force between the crate and the bed. The normal force is equal to the weight of the crate, which is mg, where m is the mass of the crate and g is the acceleration due to gravity. Therefore, the maximum acceleration is:
a = μsN/m = μsg
Substituting the given values, we get:
a = 0.400 × 9.81 m/s² = 3.92 m/s²
Therefore, the best choice is A.

Question 20

A 15-kg block is on a frictionless ramp that is inclined at 20° above the horizontal.It is connected by a very light string over an ideal pulley at the top edge of the ramp to a hanging 19-kg block,as shown in the figure.The string pulls on the 15-kg block parallel to the surface of the ramp.Find the magnitude of the acceleration of the 19-kg block after the system is gently released? A)4.0 m/s² B)3.8 m/s² C)4.2 m/s² D)4.5 m/s²

Accepted Answer

The answer of A 15-kg block is on a frictionless...

Quiz 6: Applications of Newtons Laws

A flatbed truck is carrying a 20-kg crate up a sloping road.The coefficient of static friction between the crate and the bed is 0.40,and the coefficient of kinetic friction is 0.30.What is the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place?

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?

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 on the box?

A horizontal 52-N force is needed to slide a Box 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?

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 applied to the box.What is the friction force on the box?

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?

A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N that is parallel to the slope.The coefficient of kinetic friction between the block and the slope is 0.200.What is the acceleration of the block?

An ornament of mass 40.0 g is attached to a vertical ideal spring with a force constant (spring constant) of 20.0 N/m.The ornament is then lowered very slowly until the spring stops stretching.How much does the spring stretch?

A driver in a Car traveling at 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?

You push downward on a trunk at an angle 25° below the horizontal with a force of 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?

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?

A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal.If the coefficient of kinetic friction is 0.40,at what rate does the box accelerate down the slope?

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?

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?