# A bicyclist was moving at a rate of 5 m/s, and then sped up to 9 m/s. If the cyclist has a mass of 130 kg, how much work was needed to increase his velocity?

Assuming that there are no losses, the change in kinetic energy is equal to the work done.

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The work needed to increase the velocity of the cyclist can be calculated using the work-energy principle. The change in kinetic energy is equal to the work done.

Initial kinetic energy (KE_1 = \frac{1}{2} mv_1^2)

Final kinetic energy (KE_2 = \frac{1}{2} mv_2^2)

Change in kinetic energy (ΔKE = KE_2 - KE_1)

Work done = (ΔKE)

Given:

Initial velocity (v_1 = 5 , \text{m/s})

Final velocity (v_2 = 9 , \text{m/s})

Mass of the cyclist (m = 130 , \text{kg})

Calculate (ΔKE = KE_2 - KE_1)

(KE_1 = \frac{1}{2} \times 130 \times (5)^2)

(KE_2 = \frac{1}{2} \times 130 \times (9)^2)

(ΔKE = \frac{1}{2} \times 130 \times (9)^2 - \frac{1}{2} \times 130 \times (5)^2)

(ΔKE = \frac{1}{2} \times 130 \times (81 - 25))

(ΔKE = \frac{1}{2} \times 130 \times 56)

(ΔKE = 3640 , \text{J})

Therefore, the work needed to increase the cyclist's velocity is (3640 , \text{J}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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