What is non conservatives forces and conservatives, №13583622, 05.03.2021 16:34

Physics

What conditions are required so that the mechanical energy of a system is conserved? Select all that apply. View Available Hint(s) Select all that apply. No internal forces act on objects within the system. No net external force acts on the system No non-conservative forces, such as friction, act between objects within the system

Step-by-step answer

P Answered by PhD

No non-conservative forces, such as friction, act between objects within the system

Explanation:

The law of conservation of mechanical energy in a closed system states that;'The total amount of mechanical energy, in a closed system in the absence of non-conservative forces, such as friction, remains constant.' Non-conservative forces cause energy to be lost from the system, this lost energy can't be gotten back

This implication of this is that energy can not disappear. Kinetic energy can be converted into potential energy and vice-versa in accordance with this principle.

Physics

I). mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work. ii). the principle of the conservation of mechanical energy states that the mechanical energy in a system remains constant as long as the only forces acting are conservative. iii). frictional force is a non-conservative force. which of the following statement below is false? a) for the motion of a cart on a frictionless incline plane, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential

Step-by-step answer

P Answered by Master

1

false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

mechanical energy = potential energy + kinetic energy = constant

differentiating both side

Δ potential energy + Δ kinetic energy = 0

Δ potential energy = - Δ kinetic energy

first statement is true.

Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows

change in gravitational potential energy = change in kinetic energy + work done against friction .

So given 2 nd option is incorrect.

In case of no change in gravitational energy , work done is equal to

change in kinetic energy.

Physics

In a local park, a pine cone falls off a tree branch (starting at rest) and falls to the ground. We will model the pine cone hitting the ground as if it is a mass going onto a spring (with the ground being a spring). The pine cone has a mass of 0.116 kg, and starts at a height of 7.90 m above the ground. The pine cone compresses the spring (ground) by 0.0148 m, briefly coming to rest at the bottom of its motion. (It then bounces, which is not part of the time period that we ll cover in this problem.) Assume that the pine cone, the Earth and the

Step-by-step answer

P Answered by Master

1)) ΔU = -8.96 J, 2) k = 8.18 10⁴ N / m, 3) v = 8.47 m / s

Explanation:

For this exercise we will use conservation of energy.

Starting point. Point where the pineapple comes out

Em₀ = U = m g h

where the reference frame is placed on the ground

Final point. Point where pineapple stops

Em_f = K_e + U = ½ k y² + m g y

1) the change in gravitational potential energy is

ΔU = U_f - U₀

ΔU = m g y - m g h

ΔU = mg (y-h)

let's calculate

ΔU = 0.116 9.8 (0.0148 - 7.9)

ΔU = -8.96 J

The negative sign indicates that the energy decreases

2) let's use energy conservation

Em₀ = Em_f

mg h = ½ k y² + mg y

k = mg (h-y) $\frac{2}{y^2}$

let's calculate

k = 0.116 9.8 (7.9 - 0.0148) $\frac{2}{0.0148^2}$

k = 8.18 10⁴ N / m

3) we use the same starting point and as the end point we use this height (y₂ = 4.24 m)

Em_{f2} = K + U = ½ m v² + mg y₂

energy is conserved

Em₀ = Em_{f2}

mgh = ½ m v² + m g y₂

v = $\sqrt{ 2g(h-y_2)}$

let's calculate

v = $\sqrt{ 2 \ 9.8 \ (7.9-4.24)}$

v = 8.47 m / s

Physics

What conditions are required so that the mechanical energy of a system is conserved? Select all that apply. View Available Hint(s) Select all that apply. No internal forces act on objects within the system. No net external force acts on the system No non-conservative forces, such as friction, act between objects within the system

Step-by-step answer

P Answered by PhD

No non-conservative forces, such as friction, act between objects within the system

Explanation:

The law of conservation of mechanical energy in a closed system states that;'The total amount of mechanical energy, in a closed system in the absence of non-conservative forces, such as friction, remains constant.' Non-conservative forces cause energy to be lost from the system, this lost energy can't be gotten back

This implication of this is that energy can not disappear. Kinetic energy can be converted into potential energy and vice-versa in accordance with this principle.

Physics

In a local park, a pine cone falls off a tree branch (starting at rest) and falls to the ground. We will model the pine cone hitting the ground as if it is a mass going onto a spring (with the ground being a spring). The pine cone has a mass of 0.116 kg, and starts at a height of 7.90 m above the ground. The pine cone compresses the spring (ground) by 0.0148 m, briefly coming to rest at the bottom of its motion. (It then bounces, which is not part of the time period that we ll cover in this problem.) Assume that the pine cone, the Earth and the

Step-by-step answer

P Answered by Master

1)) ΔU = -8.96 J, 2) k = 8.18 10⁴ N / m, 3) v = 8.47 m / s

Explanation:

For this exercise we will use conservation of energy.

Starting point. Point where the pineapple comes out

Em₀ = U = m g h

where the reference frame is placed on the ground

Final point. Point where pineapple stops

Em_f = K_e + U = ½ k y² + m g y

1) the change in gravitational potential energy is

ΔU = U_f - U₀

ΔU = m g y - m g h

ΔU = mg (y-h)

let's calculate

ΔU = 0.116 9.8 (0.0148 - 7.9)

ΔU = -8.96 J

The negative sign indicates that the energy decreases

2) let's use energy conservation

Em₀ = Em_f

mg h = ½ k y² + mg y

k = mg (h-y) $\frac{2}{y^2}$

let's calculate

k = 0.116 9.8 (7.9 - 0.0148) $\frac{2}{0.0148^2}$

k = 8.18 10⁴ N / m

3) we use the same starting point and as the end point we use this height (y₂ = 4.24 m)

Em_{f2} = K + U = ½ m v² + mg y₂

energy is conserved

Em₀ = Em_{f2}

mgh = ½ m v² + m g y₂

v = $\sqrt{ 2g(h-y_2)}$

let's calculate

v = $\sqrt{ 2 \ 9.8 \ (7.9-4.24)}$

v = 8.47 m / s

Physics

Block A of 15kg of the figure below slides on a surface where the coefficient of kinetic friction is uk=0.3. The block moves at 10m/s when it is s=4m from a second 10kg B block, initially at rest. Considering this situation, determine what will be the maximum compression of the spring, initially decompressed, after the collision if the coefficient of return between the blocks is e=0.6. Tip here, as the blocks are subject to friction forces, use the equation that relates mechanical energy and the work done by non-conservative forces.

Step-by-step answer

P Answered by PhD

4

0.287 m

Explanation:

The velocity of block A when it reaches block B is:

KE₀ = KE + W

½ mv₀² = ½ mv² + Fd

½ mv₀² = ½ mv² + mg μ d

v₀² = v² + 2g μ d

v² = v₀² − 2g μ d

v² = (10 m/s)² − 2 (9.8 m/s²) (0.3) (4 m)

v = 8.75 m/s

The coefficient of restitution is 0.6, so the relative velocity between A and B after the collision is:

e = |Δv after| / |Δv before|

0.6 = Δv / (8.75 m/s)

Δv = 5.25 m/s

Momentum is conserved, so the speed of block B after the collision is:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(15 kg) (8.75 m/s) + (10 kg) (0 m/s) = (15 kg) (v) + (10 kg) (v + 5.25 m/s)

131.2 m/s = 25v + 52.5 m/s

25v = 78.7 m/s

v = 3.15 m/s

Energy is conserved, so the compression of the spring is:

KE = EE + W

½ mv² = ½ kd² + Fd

½ mv² = ½ kd² + mg μ d

½ (10 kg) (3.15 m/s)² = ½ (1000 N/m) d² + (10 kg) (9.8 m/s²) (0.3) d

49.6 = 500 d² + 29.4 d

0 = 500 d² + 29.4 d − 49.6

d = [ -29.4 ± √(29.4² − 4(500)(-49.6)) ] / 1000

d = (-29.4 ± 316.2) / 1000

d = 0.287 m

Physics

Block A of 15kg of the figure below slides on a surface where the coefficient of kinetic friction is uk=0.3. The block moves at 10m/s when it is s=4m from a second 10kg B block, initially at rest. Considering this situation, determine what will be the maximum compression of the spring, initially decompressed, after the collision if the coefficient of return between the blocks is e=0.6. Tip here, as the blocks are subject to friction forces, use the equation that relates mechanical energy and the work done by non-conservative forces.

Step-by-step answer

P Answered by PhD

4

0.287 m

Explanation:

The velocity of block A when it reaches block B is:

KE₀ = KE + W

½ mv₀² = ½ mv² + Fd

½ mv₀² = ½ mv² + mg μ d

v₀² = v² + 2g μ d

v² = v₀² − 2g μ d

v² = (10 m/s)² − 2 (9.8 m/s²) (0.3) (4 m)

v = 8.75 m/s

The coefficient of restitution is 0.6, so the relative velocity between A and B after the collision is:

e = |Δv after| / |Δv before|

0.6 = Δv / (8.75 m/s)

Δv = 5.25 m/s

Momentum is conserved, so the speed of block B after the collision is:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(15 kg) (8.75 m/s) + (10 kg) (0 m/s) = (15 kg) (v) + (10 kg) (v + 5.25 m/s)

131.2 m/s = 25v + 52.5 m/s

25v = 78.7 m/s

v = 3.15 m/s

Energy is conserved, so the compression of the spring is:

KE = EE + W

½ mv² = ½ kd² + Fd

½ mv² = ½ kd² + mg μ d

½ (10 kg) (3.15 m/s)² = ½ (1000 N/m) d² + (10 kg) (9.8 m/s²) (0.3) d

49.6 = 500 d² + 29.4 d

0 = 500 d² + 29.4 d − 49.6

d = [ -29.4 ± √(29.4² − 4(500)(-49.6)) ] / 1000

d = (-29.4 ± 316.2) / 1000

d = 0.287 m

Physics

I). mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work. ii). the principle of the conservation of mechanical energy states that the mechanical energy in a system remains constant as long as the only forces acting are conservative. iii). frictional force is a non-conservative force. which of the following statement below is false? a) for the motion of a cart on a frictionless incline plane, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential

Step-by-step answer

P Answered by Specialist

1

false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

mechanical energy = potential energy + kinetic energy = constant

differentiating both side

Δ potential energy + Δ kinetic energy = 0

Δ potential energy = - Δ kinetic energy

first statement is true.

Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows

change in gravitational potential energy = change in kinetic energy + work done against friction .

So given 2 nd option is incorrect.

In case of no change in gravitational energy , work done is equal to

change in kinetic energy.

Physics

Arubber ball is dropped from rest from a height h. the ball bounces off the floor and reaches a height of 2h/3. how can we use the principle of the conservation of mechanical energy to interpret this observation? a) during the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height. b) some of the ball’s potential energy was lost in accelerating it toward the floor. c) the force of the earth’s gravity on the ball prevented it from returning to its original height. d) work was done on the ball

Step-by-step answer

P Answered by PhD

Option e.

In this case the work is done on the ball by nonconservative forces that resulted in the ball having less total mechanical energy after the bounce.

Explanation:

This is the type of nonelastic collision when a moving ball hits the ground.Although the conservation of mechanical energy possessed by the ball which is the sum of P.E and K.E., but kinetic energy is not conserved.The non conservative force did the work on the ball that after bouncing lost some of the mechanical energy of that ball. The kinetic energy in the beginning is converted in some other energy like friction and air resistance in this case.

Physics

Arubber ball is dropped from rest from a height h. the ball bounces off the floor and reaches a height of 2h/3. how can we use the principle of the conservation of mechanical energy to interpret this observation? a) during the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height. b) some of the ball’s potential energy was lost in accelerating it toward the floor. c) the force of the earth’s gravity on the ball prevented it from returning to its original height. d) work was done on the ball

Step-by-step answer

P Answered by PhD

Option e.

In this case the work is done on the ball by nonconservative forces that resulted in the ball having less total mechanical energy after the bounce.

Explanation:

This is the type of nonelastic collision when a moving ball hits the ground.Although the conservation of mechanical energy possessed by the ball which is the sum of P.E and K.E., but kinetic energy is not conserved.The non conservative force did the work on the ball that after bouncing lost some of the mechanical energy of that ball. The kinetic energy in the beginning is converted in some other energy like friction and air resistance in this case.

What is non conservatives forces and conservatives forces

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