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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

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a= 0.578 m/s

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Explanation:

Data

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m₂= 100 kg

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We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

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W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T:  cable tension : parallel to inclined plane

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W₁=m₁*g

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W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

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∑Fy = m*ay    ay = 0

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Calculated of the f

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∑Fx = m₁*ax   ,ax  =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470  -258.77 -300*a =T

T= 1211.23-300*a   Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay   ,ay  =a

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T:  cable tension:In vertical direction

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W₂=m₂*g

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∑Fy = m₂*a

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T-W₂ = m₂*a

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Problem development

Equation (1) =  Equation (2) = T

1211.23-300*a= 980  + 100*a

1211.23- 980 = 100*a + 300*a

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We replace a= 0.578 m/s in the equatión (2)

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T = 1037.8 N

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