A small object with mass m = 0.0900 kg moves along the +x-axis. The only force on the object is a conservative force that has the potential-energy function U(x)=−αx2+βx3, where α=2.00J/m2 and β=0.300J/m3. The object is released from rest at small x. When the object is at x = 4.00 m, what are its (a) speed and (b) acceleration (magnitude and direction)? (c) What is the maximum value of x reached by the object during its motion?

Answers

Answer 1

Answer:

The conservation of mechanical energy allows to find the results for the questions of the motion of mass in a conservative force are;

a) the velocity at x = 4 m is: v = 16.9 m / s

b) The acceleration at x = 4m is: a = 17.8 m / s²

c) The maximum elongations is: x = 6.67 m

Given parameters

The potential energy U = - α x² + β x³ with α= 2 j/m² and β= 0.3 j/m³

The mass of the body m = 0.0900 kg

The distance x = 4.00 m

To find

a) The speed

b) The acceleration

c) The maximum value of the distance

The conservation of mechanical energy is one of the most important concepts in physics, stable that if there is not friction the mechanical energy remains constant at all points.

Em = K + U

Where Em mechanical energy, K the kinetical energy ang U the potential energy.

a) Let's find the velocity using the conservation of mechanical energy

Starting point. Where the mass is released.

Em₀ = U (0)

Final point. When for a distance of x = 4 m here we have potential and kinetic energy.

$Em_f$ = K + U (4)

They indicate that the only force is conservative, therefore mechanicalenergy is conserved

Em₀ = $Em_f$

0 = ½ m v² + U (4)

½ m v² = -U (4)

v² = 2 / m (αx² - β x³)

Let's calculate

v² = $(2)/(0.09)$ (2 4² - 0.3 4³)

v = $√(284.44)$

v = 16.9 m / s

b) Acceleration is requested at this point.

We use that potential energy and force are related

F = $- ( dU)/(dx)$

We carry out the derivatives

F = 2αx - 3βx²

Let's calculate

F = 2 2 4 - 3 0.3 4²

F = 1.6 N

Now we use Newton's second law that relates the net force with the product of the mass and the acceleration of the body.

F = ma

a = $(F)/(m)$

a = $(1.6)/(0.09)$

a = 17.8 m / s²

c) At maximum displacement.

Let's use conservation of mechanical energy

Starting point. Where x = 0 is released

Emo = U (0) = 0

Final point. Point of maximum elongation, kinetic energy is zero

$Em_f$ = U (xmax)

Energy is conserved

Em₀ = $Em_f$

$0 = U(x_(max))$

-αx² + βx³ = 0

x² (-α + βx) = 0

the solutions of this equation is:

x = 0

-α + βx = 0

x = $(\alpha)/(\beta )$

Let's calculate

x = $(2)/(0.3)$

x = 6.67 m

In conclusion using the conservation of mechanical energy we can find the results for the questions of the motion of mass in a conservative force are;

a) the velocity at x = 4 m is: v = 16.9 m / s

b) The acceleration at x = 4m is: a = 17.8 m / s²

c) The maximum elongations is: x = 6.67 m

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Answer 2

Answer:

Answer:

a) v= 284.44 $(m)/(s^(2))$

b) a=17.78 $(m)/(s^(2) )$

c) x=6.67m

Explanation:

a).

$U(x)=-\alpha *x^(2) +\beta*x^(3)\n\alpha=2(J)/(m^(2))\n\beta=0.3(J)/(m^(2))\nU(0)+V(0)=U(4)+V(4)\nU(0)=-\alpha *0^(2) +\beta*0^(3)=0\nU(0)=0\nV(0)=0\n0=U(4)+V(4)\nU(4)=-\alpha *4^(2) +\beta*4^(3)\nU(4)=-2*4^(2)+0.3*(4^(3))\nU(4)=-12.8 J\n0=-12.8J+V(4)\n12.8=(1)/(2)*m*(v_(4))^(2) \n v_(4)^(2) =(2*12.8J)/(0.09kg)\n v_(4)^(2)=284.44$

$v_(4)=√(284.44)\nv_(4)=16.8 (m)/(s)$

b).

$F_(x)=(dU)/(dt)\nF_(x)=2*\alpha*x-3*\beta *x^(2) \nF_(4)=2*2*4-3*0.3*(4)^(2)\nF_(4)=1.6N\nF_(4)=m*a\na=(F_(4))/(m)=(1.6N)/(0.09kg)\na=17.7 (m)/(s^(2) )$

c).

$F_(x)=m*ax\nax=(F_(x))/(m) \nax=(4x-0.9x^(2))/(0.09kg)\n(dVx)/(dt)= (4x-0.9x^(2))/(0.09kg)\n\int\limits^x_x {(1)/(0.09)*(4x-0.9x^(2)) } \, dx\n (Vx^(2) )/(2)=22.2x^(2) -3.3x^(3)$

Value x,0

$Vx^(2) =44.4x^(2) -6.6x^(3)\n Vx=\sqrt{44.4x^(2) -6.6x^(3)}$

Inside the square root is the value of maximum value of x

$44.4x^(2) -6.6x^(3)=0\nx^(2)(44.4-6.6x)=0\n x=0$ but that value is not real so:

$44.4-6.6x=0\n6.6x=44.4\nx=(44.4)/(6.6)\n x=6.67m$

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The headlights are shining on a truck travelling at 100 km/h. The speed of the light from the headlights relative to the road will be:A) c

B) c + 100 km/h

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D) depends on the temperature, but faster than the speed if the truck was not moving.

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Answers

Answer:

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Speed of light is always constant irrespective of the wave source's motion and the observer's inertial frame of reference. So, no matter how fast the car is moving the speed of light will always be constant. The speed of light in air is around 299704644.54 m/s. The meter is also defined by the speed of light as 1 meter is the distance travelled by light in 1/299792458 second.

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Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.(a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.80 m/s2, what is the angular acceleration of the yo-yo in rad/s2? rad/s2.
(b) What is the angular velocity in rad/s after 0.750 s if it starts from rest? rad/s.
(c) The outside radius of the yo-yo is 3.10 cm.What is the tangential acceleration in m/s2 of a point on its edge? m/s2

Answers

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = (a)/(R)$

$\alpha = (1.80)/(0.230 * 10^(-2))$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 * 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 * 10^(-2))(782.6)$

$a_t = 24.3 m/s^2$

A disk made of many small particles of Rock and ice in orbit around a planet is called

Answers

The correct answer is ring. A disk made of many small particles of rock and ice in orbit around a planet is called ring. The rings of the planet Saturn is considered to be the most extensive planetary rings system in the Solar System.

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.9 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 70.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 40 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answers

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

a baseball bat balances 74.8 cm from one end. if an 0.542-kg glove is attached to that end, the balance point moves 25.9 cm toward the glove. calculate the mass of the bat

Answers

The mass of the bat is approximately 1.56 kg. We can use the principle of moments to solve this problem. According to the principle of moments, the sum of the moments of all the forces acting on an object is equal to zero, assuming that the object is in equilibrium.

Let's assume that the bat has a mass of m kg and its center of mass is located at a distance of x cm from the end of the bat where the glove is attached. We can then write the following equation for the moments:

mx = (m+0.542)(x+74.8-25.9)

Here, the left-hand side represents the moment of the bat about the point where the glove is attached, and the right-hand side represents the moment of the bat and the glove about the same point. We have added the distance between the center of mass of the bat and the point where the glove is attached (74.8 cm) to the distance between the glove and the point where the glove is attached (-25.9 cm) to get the total distance between the center of mass of the combined system and the point where the glove is attached.

Simplifying the equation, we get:

m*x = (m+0.542)*48.9

Expanding the brackets, we get:

mx = 48.9m + 26.56

Rearranging and solving for m, we get:

m = 0.542*74.8/(-25.9) = 1.56 kg

Therefore, the mass of the bat is approximately 1.56 kg.

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