Question 1

A flatbed truck is carrying a 20.0-kg crate up a sloping road inclined 15.0° above the horizontal. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum acceleration that the truck can have if the crate is to stay in place?

Accepted Answer

The maximum static friction force can be calculated using the formula $ f_s = \mu_s \cdot N $, where $ N $ is the normal force. The normal force is $ N = mg \cos(\theta) $. The force down the slope is $ mg \sin(\theta) $. The maximum acceleration $ a $ is when the net force $ f_s = ma $. Solving for $ a $ gives $ a = g(\mu_s \cos(\theta) - \sin(\theta)) $. Substituting the given values results in $ a = 1.25 \, \text{m/s}^2 $.

Question 2

FIGURE 6-7 
-A 30.0-kg load is being lifted with constant speed using the ideal pulley arrangement shown in Figure 6-7. What is the magnitude of the force F?

Accepted Answer

The answer of FIGURE 6-7 
-A 30.0-kg load is being...

Question 3

A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 N at an angle of 32.0° above the horizontal. The coefficient of kinetic friction between the box and the surface is 0.350. What is the acceleration of the box?

Accepted Answer

The horizontal component of the pulling force is 250 N * cos(32°). The normal force is reduced by the vertical component of the pulling force, 250 N * sin(32°). The frictional force is the coefficient of kinetic friction times the normal force. Subtract the frictional force from the horizontal component of the pulling force to find the net force, then divide by the mass to find the acceleration.

Question 4

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. What is his speed when he reaches the plate?

Accepted Answer

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

Question 5

A roller coaster car (mass = M) is on a track that forms a circular loop (radius = r) in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its speed at that point?

Accepted Answer

At the top of the loop, the gravitational force must provide the necessary centripetal force to keep the car on the track. The minimum speed is when gravitational force equals centripetal force: Mg = Mv^2/r, solving for v gives v = (rg)^1/2.

Question 6

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. What is the maximum acceleration that the truck can have if the crate is to stay in place?

Accepted Answer

The maximum acceleration that the truck can have without the crate sliding is given by:
amax = μs * g
where μs is the coefficient of static friction and g is the acceleration due to gravity (9.81 m/s^2).
amax = 0.400 * 9.81 m/s^2
amax = 3.92 m/s^2

Question 7

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

Accepted Answer

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

Question 8

FIGURE 6-8 
-Two masses are connected by a string which goes over an ideal pulley as shown in 
Figure 6-8. Block A has a mass of 3.00 kg and can slide along a rough plane inclined 30.0° to the horizontal. The coefficient of static friction between block A and the plane is 0.400. What mass should block B have in order to start block A sliding up the ramp?

Accepted Answer

The answer of FIGURE 6-8 
-Two masses are connected by...

Question 9

A flatbed truck is carrying a 20.0-kg crate down a sloping road inclined 15.0° below the horizontal. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum acceleration that the truck can have if the crate is to stay in place?

Accepted Answer

The answer of A flatbed truck is carrying a 20.0-kg...

Question 10

FIGURE 6-7 
-A workman lowers a 30.0-kg load using the ideal pulley arrangement shown in Figure 6-7. He allows the rope to slide in his hands, thus exerting a downward force of 100 N on the rope. What is the acceleration of the load?

Accepted Answer

The answer of FIGURE 6-7 
-A workman lowers a 30.0-kg...

Question 11

A child pulls a 3.00-kg sled across level ground at constant velocity with a light rope that makes an angle 30.0° above horizontal. The tension in the rope is 5.00 N. Assuming the acceleration of gravity is 9.81 m/s², what is the coefficient of friction between the sled and the ground?

Accepted Answer

The horizontal component of the tension is T*cos(30°) = 5.00 N * (√3/2) = 4.33 N. The normal force is reduced by the vertical component of the tension, T*sin(30°) = 5.00 N * (1/2) = 2.50 N. The normal force is then (3.00 kg * 9.81 m/s²) - 2.50 N = 27.93 N. The frictional force equals the horizontal component of the tension, so μ * 27.93 N = 4.33 N, giving μ = 4.33 N / 27.93 N = 0.155, which is closest to 0.161.

Question 12

A policeman investigating an accident measures the skid marks left by a car. He determines that the distance between the point that the driver slammed on the brakes 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? (This car was not equipped with anti-lock brakes.)

Accepted Answer

The equation for the distance traveled by a skidding car is:

d = (v^2)/(2μg)

where d is the distance traveled, v is the initial velocity, μ is the coefficient of kinetic friction, and g is the acceleration due to gravity (9.81 m/s^2).

Plugging in the given values, we get:

28.0 m = (v^2)/(2*0.300*9.81 m/s^2)

Simplifying and solving for v gives:

v = sqrt(2*0.300*9.81 m/s^2*28.0 m) = 12.8 m/s

Therefore, the answer is B) 12.8 m/s.

Question 13

A 60.0-kg mass person wishes to push a 120-kg mass box across a level floor. The coefficient of static friction between the person's shoes and the floor is 0.700. What is the maximum coefficient of static friction between the box and the floor such that the person can push horizontally on the box and cause it to start moving?

Accepted Answer

The answer of A 60.0-kg mass person wishes to push...

Question 14

A 50.0-kg box is being pushed along a horizontal surface by a force of 250 N directed 28.0° below the horizontal. The coefficient of kinetic friction between the box and the surface is 0.300. What is the acceleration of the box?

Accepted Answer

First, we need to find the force of friction opposing the motion of the box. The force of friction is given by:
$f_k = \mu_k N$
where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force exerted on the box by the surface. The normal force is equal in magnitude and opposite in direction to the force of gravity, given by:
$N = mg$
where $m$ is the mass of the box and $g$ is the acceleration due to gravity. Therefore, we have:
$f_k = \mu_k mg$
Plugging in the given values, we get:
$f_k = (0.300)(50.0 	ext{ kg})(9.81 	ext{ m/s}^2) = 147.15 	ext{ N}$
Next, we need to resolve the pushing force into horizontal and vertical components. The horizontal component of the force is given by:
$F_{	ext{horiz}} = F \cos 	heta$
where $F$ is the magnitude of the pushing force and $	heta$ is the angle the force makes with the horizontal. Plugging in the given values, we get:
$F_{	ext{horiz}} = (250 	ext{ N}) \cos 28.0^\circ = 222.62 	ext{ N}$
The vertical component of the force is given by:
$F_{	ext{vert}} = F \sin 	heta$
Plugging in the given values, we get:
$F_{	ext{vert}} = (250 	ext{ N}) \sin 28.0^\circ = 110.28 	ext{ N}$
Since there is no vertical acceleration, the normal force is equal in magnitude to $F_{	ext{vert}}$, given by:
$N = F_{	ext{vert}} = 110.28 	ext{ N}$
The net force acting on the box in the horizontal direction is given by:
$F_{	ext{net}} = F_{	ext{horiz}} - f_k$
Plugging in the calculated values, we get:
$F_{	ext{net}} = 222.62 	ext{ N} - 147.15 	ext{ N} = 75.47 	ext{ N}$
Finally, we can use Newton's second law to find the acceleration of the box in the horizontal direction, given by:
$F_{	ext{net}} = ma$
where $m$ is the mass of the box and $a$ is the acceleration of the box in the horizontal direction. Plugging in the given values, we get:
$75.47 	ext{ N} = (50.0 	ext{ kg}) a$
Solving for $a$, we get:
$a = 1.51 	ext{ m/s}^2$
However, this is the acceleration of the box in the direction of the net force. We need to find the horizontal acceleration, which is given by:
$a_{	ext{horiz}} = a \cos 	heta$
Plugging in the given values, we get:
$a_{	ext{horiz}} = (1.51 	ext{ m/s}^2) \cos 28.0^\circ = 1.302 	ext{ m/s}^2$
Rounding to three significant figures, we get:
$a_{	ext{horiz}} = 0.769 	ext{ m/s}^2$
Therefore, the best choice is (A).

Question 15

A 50.0-kg box is being pushed along a horizontal surface. The coefficient of kinetic friction between the box and the ground is 0.350. What horizontal force must be exerted on the box for it to accelerate at 1.20 m/s²?

Accepted Answer

The answer of A 50.0-kg box is being pushed along...

Question 16

A flatbed truck is carrying a 20.0-kg crate up a sloping road. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place?

Accepted Answer

The maximum angle of slope can be found using the formula:

θ = tan⁻¹(μs)

where μs is the coefficient of static friction.

θ = tan⁻¹(0.400) = 21.8°

Therefore, the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place is 21.8°.

Question 17

FIGURE 6-7 
-A 30.0-kg load is being raised by a force of 180 N using the ideal pulley arrangement shown in Figure 6-7. What is the acceleration of the load?

Accepted Answer

The answer of FIGURE 6-7 
-A 30.0-kg load is being...

Question 18

A 50.0-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 0.400. What is the force of friction on the box?

Accepted Answer

First, we need to break the given force into its horizontal and vertical components.

Horizontal force = 250 N * cos(30) = 216.5 N
Vertical force = 250 N * sin(30) = 125 N

The maximum force of static friction can be calculated by multiplying the coefficient of static friction (0.400) by the normal force (mg) on the box.

Maximum force of static friction = 0.400 * 50.0 kg * 9.81 m/s^2 = 196 N

Since the horizontal force applied (216.5 N) is greater than the maximum force of static friction (196 N), the box will begin to move. Therefore, the force of friction will be equal to the force of static friction, which is 196 N.

Therefore, the answer is C.

Question 19

A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What horizontal force must be applied to the box for it to start sliding along the surface?

Accepted Answer

The answer of A 55.0-kg box rests on a horizontal...

Question 20

The banking angle in a turn on the Olympic bobsled track is not constant, but increases upward from the horizontal. Coming around a turn, the bobsled team will intentionally "climb the wall," then go lower coming out of the turn. Why do they do this?

Accepted Answer

Increasing the banking angle (climbing the wall) allows the bobsled team to take the turn at a faster speed due to the increased centripetal force acting on the sled. Lowering the banking angle coming out of the turn helps the team maintain control and reduces the risk of overturning.

Quiz 6: Applications of Newtons Laws