A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

Hot temperature TH= 300K

Cold temperature TL= 260K

Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

Wnet, in = QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

Thus, the 133.33J represents the initial work input during the heat transfer process.

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

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Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.