Ask question

Ask question

All categories

3 years ago

13

Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t

he car on the curve is 16 m/s, but the box remains stationary relative to the floor of the trunk. Determine the minimum coefficient of static friction for the box on the floor of the trunk.

1 answer:

LiRa [457]3 years ago

3 0

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

$F_{c} = \frac{mv^{2}}{R}$

$f_{s} = \mu_{s}mg$

where

$\mu_{s}$ = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

$F_{net} = f_{s} - F_{c}$

$0 = f_{s} - F_{c}$

$\mu_{s}mg = \frac{mv^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{gR}$

$\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544$

You might be interested in

)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app

masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

$F \propto \frac{1}{r}$

$F = \frac{k}{r}$ (Eq. 1)

Where:

$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

Where:

$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

$F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)$

$F_{B} = 12.857\,N$

A force of 12.857 newtons must be applied to open the door.

3 0

2 years ago

Which is the equivalent resistance of the circuit<br><br> shown below?

Misha Larkins [42]

1/Rt=1/R1+1/R2+1/R3, 1/Rt=1/3+1/12+1/4=2/3, Rt=equivalent resistance= 1.5 ohms

8 0

3 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

Umnica [9.8K]

Answer:

<em>A = 6.9 cm</em>

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

$\displaystyle w=\sqrt{\frac{k}{m}}$

The equation for the motion of the object is written as a sinusoid:

$\displaystyle X=A\ cos\ w\ t$

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

$\displaystyle X'=V=-A\ w\ sin\ w\ t$

And the maximum speed is

$\displaystyle V_{max}= A\ w$

Solving for the amplitude

$\displaystyle A= \frac{V_{max}}{w}$

Computing w

$\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\ rad/ s$

Calculating A

$\displaystyle A=\frac{1.7}{24.5}=0.069\ m$

$\displaystyle \boxed{A=6.9\ cm}$

7 0

3 years ago

Read the scenario and solve these two problems.

Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy $K$ of an object is given by:

$K=\frac{1}{2}mV^{2}$

Where:

$m=12000 kg$ is the mass of the train

$V=30 m/s$ is the speed of the train

Solving the equation:

$K=\frac{1}{2}(12000 kg)(30 m/s)^{2}$

$K=5400000 J$ This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

$E_{i}=E_{f}$

$K_{i}+P_{i}=K_{f}+P_{f}$

Where:

$K_{i}=5400000 J$ is the train's initial kinetic energy

$P_{i}=0 J$ is the train's initial potential energy

$K_{f}=0 J$ is the train's final kinetic energy

$P_{f}=mgh$ is the train's final potential energy, where $g=9.8 m/s^{2}$ is the acceleration due gravity and $h$ is the height.

Rewriting the equation with the given values:

$5400000 J=(12000 kg)(9.8 m/s^{2})h$

Finding $h$ :

$h=45.918 m \approx 45.92 m$

7 0

2 years ago

Read 2 more answers

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0

2 years ago