1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ehidna [41]
3 years ago
8

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N

/m; the other end of the spring is fixed in place. The cookie has a kinetic energy of 20.0 J as it passes through the spring's equilibrium position. As the cookie slides, a frictional force of magnitude 11.0 N acts on it. (a) How far will the cookie slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the cookie as it slides back through the equilibrium position?
Physics
1 answer:
irinina [24]3 years ago
3 0

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

You might be interested in
Suppose a car of mass m is moving at a constant speed v of
SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

\arcsin(v^2/(gR))

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

w_r = mg\sin(\theta)

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g

Also, the car is making a circular motion:

a = \frac{v^2}{R}

Combining the equations:

\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}

Finally the angle is

\arcsin(v^2/(gR))

4 0
3 years ago
A 9,100​-microfarad ​[mu​F] capacitor is charged to 22 volts​ [V]. If the capacitor is completely discharged through an iron rod
MakcuM [25]

Answer:

The temperature rises by 0.52K

Explanation:

Detailed explanation and calculation is shown in the image below

5 0
3 years ago
A snail travels 300 cm in 4 minutes.calculate speed of snail in m/s
LekaFEV [45]

Answer:

\frac{0.0125m}{s}

Explanation:

In order to solve this question we need to know that  speed = \frac{distance}{time}. Then we need to convert 4 minutes into seconds and cm into m. We can do that by multiplying the number of minutes by 60 (because there is 60 seconds in one minute) and dividing the number of cm by 100 (because there is 100 cm in one m). So.......

4min = 4 x 60s = 240s

300cm = 300/100 m = 3m

Now we know that distance = 300m, and that the time = 4min = 240s ⇒

⇒ speed=\frac{distance}{time} = \frac{3m}{240s} = \frac{0.0125m}{s}

5 0
3 years ago
The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Dril
STatiana [176]

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

                         T = 5252*P / N

                         T = 5252*5.85 / 1900

                         T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

                          T = F*r

Where,   r = d / 2

                          F = 2T / d

                          F = 2*16.171 / 0.98

                          F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

4 0
3 years ago
Read 2 more answers
At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of
photoshop1234 [79]

The recoil velocity of cannon is (4) 5.0 m/s

Explanation:

We can find the recoil velocity from the law of conservation of momentum.

The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.

Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec

m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.

So recoil velocity of cannon v2 is given by,

v2 = -(m1÷m2)v1

v2 = -(100÷500)15

v2 = -5 m/s

where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.

Hence, option (4)5.0 m/s is the correct answer.

6 0
3 years ago
Read 2 more answers
Other questions:
  • Explain how airplane wings and the wings of birds affect the air flowing over them to produce lift
    15·1 answer
  • ________ can occur when an analog connection creates an electromagnetic field around its conductors, inducing its waveforms on a
    7·1 answer
  • Select the statement that correctly completes the description of phase difference.
    5·1 answer
  • Describe how all the air in the oven becomes hot?
    6·1 answer
  • A 25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the left and from the end of a se
    5·2 answers
  • Why would it be hard to find the ideal light intensity if the temperure were very hot or cold?
    7·1 answer
  • What is the term that describes<br> the material through which a wave travels?
    7·1 answer
  • The shaded boxes contain the first half of four statements. The unshaded boxes
    15·1 answer
  • A 50 N girl climbs the flight of stairs in 3 seconds. How much work does she
    11·1 answer
  • Help me I don't know what I'm doing ​
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!