Question 1

The figure shows a 2000 kg cable car descending a high hill. A counterweight of mass 1800 kg on the other side of the hill aids the brakes in controlling the cable car's speed. The rolling friction of both the cable car and the counterweight are negligible. How much braking force does the cable car need to descend at constant speed? 
A) 3800 N
B) 2900 N
C) 2000 N
D) 980 N

Accepted Answer

The answer of The figure shows a 2000 kg cable...

Question 2

A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be traveling when it tires just barely lose contact with the road when the car is at the top of the hill?
A) 45.5 m/s
B) 41.8 m/s
C) 34.3 m/s
D) 22.2 m/s
E) 27.6 m/s

Accepted Answer

When the car reaches the top of the hill, the normal force acting on the car would be zero as the weight will be perpendicular to the road. Thus, the weight of the car should be equal to the centrifugal force acting on the car given by:

mv^2/r = mg

where m = mass of the car, v = velocity of the car, r = radius of curvature, and g = acceleration due to gravity.

Rearranging the equation, we get:

v = √(gr)

Substituting the values, we get:

v = √(9.81 m/s^2 × 0.120 km × 1000 m/km) ≈ 34.3 m/s

Therefore, the car would be traveling at a speed of 34.3 m/s or close to 77 mph when its tires just barely lose contact with the road. Hence, the best choice is C.

Question 3

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coefficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?
A) 1590 N
B) 3430 N
C) 7240 N
D) 7820 N
E) 795 N

Accepted Answer

The correct answer is A because the magnitude of the force exerted by friction on the car is 1590 N. This is because the force of friction is equal to the coefficient of static friction times the normal force, which is equal to the weight of the car minus the component of the normal force that is perpendicular to the road.

Question 4

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?
A) 1.23
B) 0.816
C) 0.736
D) 0.952
E) 0.662

Accepted Answer

The minimum coefficient of static friction ($μ_s$) required to prevent the car from sliding can be found using the formula for centripetal force ($F_c$) required to keep an object moving in a circle of radius $r$ at a constant speed $v$, which is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the object. The frictional force ($F_f$) that provides this centripetal force is given by $F_f = μ_sN$, where $N$ is the normal force. In this case, since the car is on a horizontal surface, $N = mg$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). Setting $F_c = F_f$, we get $\frac{mv^2}{r} = μ_smg$. The mass ($m$) cancels out, and solving for $μ_s$ gives $μ_s = \frac{v^2}{rg}$. Substituting the given values ($v = 40.0 m/s$, $r = 200 m$, and $g = 9.8 m/s^2$) into the equation yields $μ_s = \frac{(40.0)^2}{200 \times 9.8} = \frac{1600}{1960} = 0.816$, which matches choice B.

Question 5

A system of blocks and a frictionless pulley is shown in the figure. Block A has a mass of 2.0 kg and is on a rough horizontal surface for which μ_s = 0.40 between the surface and block A. The rope pulls horizontally on block A. Block C has a mass of 1.0 kg. An external force P = 23.0 N, applied vertically to block A, maintains the system at rest as shown in the figure. What is the friction force on block A? A) 6.3 N B) 6.9 N C) 7.5 N D) 5.7 N E) 5.1 N

Accepted Answer

The answer of A system of blocks and a frictionless...

Question 6

The figure shows two forces, each of magnitude 4.6 N, acting on an object. The angle between these forces is 40°, and they make equal angles above and below the horizontal. What third force will cause the object to be in equilibrium (acceleration equals zero)? 
A) 8.6 N pointing to the right
B) 7.0 N pointing to the right
C) 4.3 N pointing to the right
D) 3.5 N pointing to the right

Accepted Answer

The answer of The figure shows two forces, each of...

Question 7

Two blocks are connected by a string that goes over an ideal pulley as shown in the figure and pulls on block A parallel to the surface of the plane. 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 plane?

Accepted Answer

The answer of Two blocks are connected by a string...

Question 8

Engineers are designing a curved section of a highway. If the radius of curvature of the curve is 194 m, at what angle should the curve be banked so that a car traveling at 29.0 m/s will stay on the road without the aid of frictional forces?
A) 23.9°
B) 16.7°
C) 21.0°
D) 26.1°

Accepted Answer

The banking angle θ can be calculated using the formula tan(θ) = v²/(r*g), where v is the velocity, r is the radius, and g is the acceleration due to gravity (approximately 9.81 m/s²). Substituting the given values, tan(θ) = (29.0)² / (194 * 9.81), which gives θ ≈ 23.9°.

Question 9

A 23 kg mass is connected to a nail on a frictionless table by a massless string 1.3 m long. There is no appreciable friction between the nail and the string. If the tension in the string is 51 N while the mass moves in a uniform circle on the table, how long does it take for the mass to make one complete revolution?
A) 4.8 s
B) 3.8 s
C) 4.5 s
D) 5.2 s

Accepted Answer

The correct answer is A.Explanation:The tension in the string provides the centripetal force necessary to keep the mass moving in a circle. The centripetal force is given by:$$F_c = mv^2/r$$where:* Fc is the centripetal force* m is the mass* v is the speed* r is the radius of the circleThe tension in the string is equal to the centripetal force, so:$$T = mv^2/r$$We can solve this equation for v:$$v = sqrt(Tr/m)$$The period of revolution is the time it takes for the mass to make one complete revolution. The period is given by:$$T = 2πr/v$$We can substitute the expression for v into this equation:$$T = 2πr sqrt(m/T)$$We can solve this equation for T:$$T = 2π sqrt(mr/T)$$We can plug in the given values:$$T = 2π sqrt(23 kg * 1.3 m / 51 N)$$$$T = 4.8 s$$Therefore, it takes 4.8 s for the mass to make one complete revolution.

Question 10

A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the road is 0.600. What is the maximum speed at which the car can travel around this curve without sliding?
A) 29.6 m/s
B) 33.1 m/s
C) 24.8 m/s
D) 42.0 m/s
E) 37.9 m/s

Accepted Answer

The answer of A car enters a 300-m radius horizontal...

Question 11

Block A of mass 8.0 kg and block X are attached to a rope that passes over a pulley. A 50-N force P is applied horizontally to block A, keeping it in contact with a rough vertical face. The coefficients of static and kinetic friction between the wall and block A are μ_s = 0.40 and μ_k = 0.30. The pulley is light and frictionless. In the figure, the mass of block X is adjusted until block A descends at constant velocity of 4.75 cm/s when it is set into motion. What is the mass of block X? A) 6.5 kg B) 7.2 kg C) 8.0 kg D) 8.8 kg E) 9.5 kg

Accepted Answer

The answer of Block A of mass 8.0 kg and...

Question 12

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static friction between the tires and the road is 0.350. At what speed will the car begin to skid sideways?
A) 9.25 m/s
B) 23.6 m/s
C) 34.3 m/s
D) 35.0 m/s
E) 18.5 m/s

Accepted Answer

The car will begin to skid sideways when the centripetal force required to keep the car moving in a circle exceeds the maximum static friction force. The maximum static friction force can be calculated using $f_{\text{max}} = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. For a car of mass $m$ on a horizontal surface, $N = mg$, where $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$). The centripetal force required for circular motion is given by $F_c = \frac{mv^2}{r}$, where $v$ is the velocity of the car and $r$ is the radius of the curve. Setting the maximum static friction force equal to the required centripetal force gives $\mu_s mg = \frac{mv^2}{r}$. Solving for $v$ gives $v = \sqrt{\mu_s g r}$. Substituting $\mu_s = 0.350$, $g = 9.8 \, \text{m/s}^2$, and $r = 100 \, \text{m}$ gives $v = \sqrt{0.350 \times 9.8 \times 100} = \sqrt{343} \approx 18.5 \, \text{m/s}$.

Question 13

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate and ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate? A) 8.4 m/s² B) 10 m/s² C) 12 m/s² D) 13 m/s² E) 3.4 m/s²

Accepted Answer

To find the acceleration, calculate the net force acting on the crate. The horizontal force has a component along the ramp, and friction and gravity oppose the motion. Use the formula: net force = applied force component - friction force - gravitational force component. Then, use F = ma to find acceleration. The correct calculation gives approximately 8.4 m/s².

Question 14

The figure shows two forces acting at right angles on an object. They have magnitudes F₁ = 6.3 N and F₂ = 2.1 N. What third force will cause the object to be in equilibrium (acceleration equals zero)? A) 6.6 N at 162° counterclockwise from ₁ B) 6.6 N at 108° counterclockwise from ₁ C) 4.2 N at 162° counterclockwise from ₁ D) 4.2 N at 108° counterclockwise from ₁

Accepted Answer

The answer of The figure shows two forces acting at...

Question 15

A 600-kg car traveling at 24.5 m/s is going around a curve having a radius of 120 m that is banked at an angle of 20°.
(a) Is the curve properly banked for the car's speed?
(b) What is the minimum coefficient of static friction required between the road and the car's tires so the car does not skid?

Accepted Answer

The answer of A 600-kg car traveling at 24.5 m/s...

Question 16

A traffic light weighing 100 N is supported by two ropes as shown in the figure. The tensions in the ropes are closest to 
A) 50 N.
B) 56 N.
C) 63 N.
D) 66 N.
E) 83 N.

Accepted Answer

The answer of A traffic light weighing 100 N is...

Question 17

A 1520-N crate is to be held in place on a ramp that rises at 30.0° above the horizontal (see figure). The massless rope attached to the crate makes a 22.0° angle above the surface of the ramp. The coefficients of friction between the crate and the surface of the ramp are µ_k = 0.450 and µ_s = 0.650. The pulley has no appreciable mass or friction. What is the MAXIMUM weight w that can be used to hold this crate stationary on the ramp?

Accepted Answer

The answer of A 1520-N crate is to be held...

Question 18

In the figure, a block of mass M hangs at rest. The rope that is fastened to the wall is horizontal and has a tension off 52 N. The rope that is fastened to the ceiling has a tension of 91 N, and makes an angle θ with the ceiling. What is the angle θ? 
A) 55°
B) 35°
C) 30°
D) 63°
E) 45°

Accepted Answer

We can start by drawing a free body diagram of the block:

```
    T1
     ↑
     │
     M
     ↓
T2 ←–––––––┘
```

where `T1` is the tension in the horizontal rope and `T2` is the tension in the diagonal rope. We know that the block is at rest, so the net force on it must be zero. In other words, the magnitudes of the two tensions must be equal:

```
T1 = T2
```

We also know that `T1 = 52 N` and `T2 = 91 N`. So we can substitute and solve for the angle `θ`:

```
T2 sin(θ) = T1
91 sin(θ) = 52
sin(θ) = 52/91
θ ≈ 55°
```

Therefore, the answer is A) 55°.

Question 19

A 600-kg car is going around a banked curve with a radius of 110 m at a speed of 24.5 m/s. What is the appropriate banking angle so that the car stays on its path without the assistance of friction?
A) 29.1°
B) 13.5°
C) 33.8°
D) 56.2°
E) 60.9°

Accepted Answer

The banking angle θ can be found using the formula tan(θ) = v²/(r*g), where v is the speed, r is the radius, and g is the acceleration due to gravity (9.8 m/s²). Substituting the given values, tan(θ) = (24.5)²/(110*9.8), which gives θ ≈ 29.1°.

Question 20

Two weights are connected by a massless wire and pulled upward with a constant speed of 1.50 m/s by a vertical pull P. The tension in the wire is T (see figure). P is closest to 
A) 25 N.
B) 125 N.
C) 187.5 N.
D) 245 N.
E) 1225 N.

Accepted Answer

The answer of Two weights are connected by a massless...

Quiz 5: Applying Newtons Laws

A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be traveling when it tires just barely lose contact with the road when the car is at the top of the hill?

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coefficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?

The figure shows two forces, each of magnitude 4.6 N, acting on an object. The angle between these forces is 40°, and they make equal angles above and below the horizontal. What third force will cause the object to be in equilibrium (acceleration equals zero) ?

Engineers are designing a curved section of a highway. If the radius of curvature of the curve is 194 m, at what angle should the curve be banked so that a car traveling at 29.0 m/s will stay on the road without the aid of frictional forces?

A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the road is 0.600. What is the maximum speed at which the car can travel around this curve without sliding?

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static friction between the tires and the road is 0.350. At what speed will the car begin to skid sideways?

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate and ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate?

The figure shows two forces acting at right angles on an object. They have magnitudes F₁ = 6.3 N and F₂ = 2.1 N. What third force will cause the object to be in equilibrium (acceleration equals zero) ?

A 600-kg car traveling at 24.5 m/s is going around a curve having a radius of 120 m that is banked at an angle of 20°. (a) Is the curve properly banked for the car's speed? (b) What is the minimum coefficient of static friction required between the road and the car's tires so the car does not skid?

A traffic light weighing 100 N is supported by two ropes as shown in the figure. The tensions in the ropes are closest to

In the figure, a block of mass M hangs at rest. The rope that is fastened to the wall is horizontal and has a tension off 52 N. The rope that is fastened to the ceiling has a tension of 91 N, and makes an angle θ with the ceiling. What is the angle θ?

A 600-kg car is going around a banked curve with a radius of 110 m at a speed of 24.5 m/s. What is the appropriate banking angle so that the car stays on its path without the assistance of friction?

Two weights are connected by a massless wire and pulled upward with a constant speed of 1.50 m/s by a vertical pull P. The tension in the wire is T (see figure) . P is closest to