Question 1

Diff: 1 Var: 5 Page Ref: Sec. 6-2
-A mass of 40.0 grams is attached to a vertical spring with a spring constant k = 20.0 N/m and lowered slowly until the spring stops stretching. How much does the spring stretch?

Accepted Answer

The formula for the displacement of a spring is given by:

x = F/k

where:
x = displacement
F = force applied
k = spring constant

Using the given information, we can find the force applied to the spring:

F = mg = (0.040 kg)(9.81 m/s²) = 0.392 N

Substituting this into the formula for x, we get:

x = F/k = (0.392 N)/(20.0 N/m) = 0.0196 m

Therefore, the spring stretches by 0.0196 m. The best choice is B.

Question 2

FIGURE 6-11 
-Refer to Figure 6-11. Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. What is the acceleration of the masses?

Accepted Answer

Since there is no friction between block A and the table, the tension in the string is equal on both sides of the pulley. Applying Newton's second law to block A, we get $\sum F = ma_A = T$. For block B, we have $\sum F = ma_B = mg - T$, where $g$ is the acceleration due to gravity. Since both blocks have the same acceleration, we can eliminate $T$ between these two equations to get $a = g\frac{m_A - m_B}{m_A + m_B}$. Plugging in the given values, we get $a = 9.81 \frac{3.00-2.00}{3.00+2.00} \approx 3.92 	ext{ m/s}^2$. Therefore, the correct answer is choice C.

Question 3

A 50.0-kg block is being pushed up a 15.0° slope by a force of 300 N which is directed 30.0° below the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the acceleration of the block?

Accepted Answer

First, we need to resolve the force of 300 N into its components parallel and perpendicular to the slope. The force parallel to the slope is $300\sin 30^\circ = 150$ N, and the force perpendicular to the slope is $300\cos 30^\circ = 259.81$ N.

Next, we need to find the force of kinetic friction acting on the block, which is $\mu_k N$, where $N$ is the normal force. The normal force is equal to the perpendicular force, which is $259.81$ N.

Therefore, the force of kinetic friction is $\mu_k N=0.200(259.81)=51.96$ N.

Now, we can apply Newton's second law along the direction parallel to the slope:
$$\sum F = ma$$
$$150-51.96=50.0a\sin 15^\circ$$
Solving for $a$, we get:
$$a=0.164	ext{ m/s}^2$$

Therefore, the correct choice is A.

Question 4

FIGURE 6-9 
-A 22.0-kg crate is pulled along a horizontal floor by the ideal arrangement shown in 
Figure 6-9. The force F is 300 N. The coefficient of friction between the crate and the floor is 0.270. What is the acceleration of the crate?

Accepted Answer

The net force is the applied force minus the frictional force. Frictional force is calculated as the product of the normal force and the coefficient of friction. The normal force is equal to the weight of the crate (mass times gravity). Using Newton's second law, the acceleration is the net force divided by the mass.

Question 5

FIGURE 6-10 
-A 10-kg sign is held by two ropes as shown in Figure 6-10. What is the tension on rope A?

Accepted Answer

The answer of FIGURE 6-10 
-A 10-kg sign is held...

Question 6

A 3.00-kg mass rests on the ground. It is attached to a string which goes vertically to and over an ideal pulley. A second mass is attached to the other end of the string and released. The 3.00-kg mass rises 50.0 cm in 1.00 s. How large was the other mass?

Accepted Answer

The tension in the string is equal to the weight of the 3.00-kg mass, which is 29.4 N. The acceleration of the system is therefore 29.4 N / (3.00 kg + m2) = 9.80 m/s2. The distance traveled by the 3.00-kg mass in 1.00 s is 0.500 m, so the acceleration is 9.80 m/s2. Therefore, the mass of the other mass is 3.67 kg.

Question 7

FIGURE 6-11 
-Refer to Figure 6-11. Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. How long does it take block B to travel 80.0 cm?

Accepted Answer

The acceleration of the system is given by $a=\frac{m_2g-m_1g}{m_1+m_2}$. From the kinematic equation $x=ut+\frac{1}{2}at^2$, where $x=80\ 	ext{cm}$, $u=0$, and $a$ is found above, we can solve for $t$. Converting cm to m and plugging in the given values, we get $t = \sqrt{\frac{2x}{a}} = \sqrt{\frac{2(0.80)}{\frac{2}{5}(3+2)}} = 0.639\ 	ext{s}$, which gives us the correct answer of choice C.

Question 8

A 50.0-kg block is being pulled up a 15.0° slope by a force of 300 N which is directed 30.0° above the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the acceleration of the block?

Accepted Answer

The correct answer is A because the acceleration of the block is 1.36 m/s². The force of friction is 98.0 N, and the net force on the block is 202 N. The mass of the block is 50.0 kg, so the acceleration is 202 N / 50.0 kg = 1.36 m/s².

Question 9

A 3.00-kg mass and a 5.00-kg mass hang vertically at the ends of a rope that goes over an ideal pulley. If the masses are released from rest, how long does it take for the 3.00-kg mass to rise by 1.00 m?

Accepted Answer

The answer of A 3.00-kg mass and a 5.00-kg mass...

Question 10

A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N which 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?

Accepted Answer

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

Question 11

FIGURE 6-12 
-Two masses are connected by a string which goes over an ideal pulley as shown in Figure 6-12. Block A has a mass of 3.0 kg and can slide along a smooth plane inclined 30° to the horizontal. What is the mass of block B if the system is in equilibrium?

Accepted Answer

The tension in the string is equal to the weight of block B. The component of the weight of block A parallel to the plane is equal to the tension in the string. Therefore, the mass of block B is equal to the mass of block A times the sine of the angle of the plane.$$m_B = m_A \sin 	heta$$$$m_B = 3.0 	ext{ kg} \sin 30° = 1.5 	ext{ kg}$$

Question 12

A 50.0-kg block is being pulled up a 13.0° slope by a force of 250 N which is parallel to the slope, but the block does not slide up the slope. What is the minimum value of the coefficient of static friction required for this to happen?

Accepted Answer

The force of static friction must be equal to the component of gravitational force pulling the block down the slope minus the applied force. The normal force is the component of gravitational force perpendicular to the slope. Using these, the minimum coefficient of static friction can be calculated as 0.292.

Question 13

A 3.00-kg mass and a 5.00-kg mass hang vertically at the ends of a rope that goes over an ideal pulley. If the masses are released, what is the resulting acceleration of the masses?

Accepted Answer

The answer of A 3.00-kg mass and a 5.00-kg mass...

Question 14

A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring attached to a wall. The spring is aligned with the plane and has a spring constant of 120 N/m. How much does the spring stretch?

Accepted Answer

The answer of A mass of 3.0 kg rests on...

Question 15

A box slides down a 25.0° slope under its own weight. The coefficient of kinetic friction between the box and the slope is 0.350. What is the acceleration of the box?

Accepted Answer

The acceleration of the box is given by:$$a = g(sinθ - μkcosθ)$$where:* g is the acceleration due to gravity (9.8 m/s^2)* θ is the angle of the slope (25.0°)* μk is the coefficient of kinetic friction (0.350)Plugging in the values, we get:$$a = 9.8 m/s^2(sin25.0° - 0.350cos25.0°) = 1.03 m/s^2$$Therefore, the acceleration of the box is 1.03 m/s^2.

Question 16

A tightrope walker with a mass of 60.0 kg stands at the center of a rope which was initially strung horizontally between two poles. His weight causes the rope to sag symmetrically, making an angle of 4.80° with the horizontal. What is the tension in the rope?

Accepted Answer

The answer of A tightrope walker with a mass of...

Question 17

FIGURE 6-10 
-A 10-kg sign is held by two ropes as shown in Figure 6-10. What is the tension on rope B?

Accepted Answer

The sign is in equilibrium, meaning that the net force acting on it is zero. Taking the sum of the forces along the y-axis, we get:
T_Acos(30°) + T_Bcos(45°) -mg = 0
where T_A and T_B are the tensions on ropes A and B, respectively, and m is the mass of the sign. Plugging in the known values, we get:
T_A(√3/2) + T_B(√2/2) - 10 kg × 9.8 m/s² = 0
Since T_A = 50 N (given in the figure), we can solve for T_B:
T_B = (10 kg × 9.8 m/s² - T_A(√3/2)) / (√2/2) ≈ 72 N

Question 18

FIGURE 6-11 
-Refer to Figure 6-11. Block A has a mass of 2.00 kg and rests on a rough table and is connected to block B, which has a mass of 3.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. The coefficient of kinetic friction between block A and the table is 0.300. What is the acceleration of the masses?

Accepted Answer

The answer of FIGURE 6-11 
-Refer to Figure 6-11. Block...

Question 19

A 50.0-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 26.0° below the horizontal is applied to the box. What is the minimum coefficient of static friction between the box and the surface required for the box to remain stationary?

Accepted Answer

First, we need to find the normal force acting on the box. This force is equal to the weight of the box, which is (50.0 kg)(9.81 m/s^2) = 490.5 N.

Next, we need to find the force of friction required to keep the box from sliding. Using trigonometry, we can find that the horizontal component of the applied force is (250 N)cos(26.0°) = 221.3 N. Therefore, the force of friction must also be 221.3 N in the opposite direction to balance out the horizontal forces.

Finally, we can use the formula for static friction to find the minimum coefficient of static friction required:

f_static = μ_static * N

Where f_static is the force of friction, N is the normal force, and μ_static is the coefficient of static friction. Plugging in our values, we get:

221.3 N = μ_static * (490.5 N)

μ_static = 0.451

This value is closest to choice C, 0.375, so that is our answer.

Question 20

A 50.0-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 20.7° below the horizontal is applied to the box. The coefficient of kinetic friction between the box and the surface is 0.300. What is the acceleration of the box?

Accepted Answer

The horizontal component of the applied force is 250 N * cos(20.7°). The normal force is increased by the vertical component of the applied force, 250 N * sin(20.7°), and the weight of the box. The frictional force is the coefficient of kinetic friction times the normal force. Subtract the frictional force from the horizontal component of the applied force to find the net force, then use F = ma to find the acceleration.

Quiz 6: Applications of Newtons Laws

Diff: 1 Var: 5 Page Ref: Sec. 6-2 -A mass of 40.0 grams is attached to a vertical spring with a spring constant k = 20.0 N/m and lowered slowly until the spring stops stretching. How much does the spring stretch?

FIGURE 6-11 -Refer to Figure 6-11. Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. What is the acceleration of the masses?