Get the answer to "Determine the energy required to accelerate a 1300-kg..." - Plainmath

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Russell Gillen

Answered question

2021-12-19

Determine the energy required to accelerate a 1300-kg car from 10 to 60 km/h on an uphill road with a vertical rise of 40 m.

Answer & Explanation

peterpan7117i

Beginner2021-12-20Added 39 answers

Givens:

m = 1300 k g

V_{1} = 10 k \frac{m}{h}

V_{2} = 60 k \frac{m}{h} V

Δ z = 40 m

g = 9.81 \frac{m}{s^{2}}

Solution:
Note :

1 k \frac{m}{h} = \frac{1000 m}{60 \times 60 s} = \frac{5}{18}; \frac{m}{s}

The required energy is the sum of the change in potential energy and the change in kinetic energy.

Δ E = m g Δ z + m \frac{Δ V^{2}}{2} = m g Δ z + m \frac{V_{2}^{2} - V_{1}^{2}}{2} = 1300 [9.81 \times 40 + \frac{60^{2} - 10^{2}}{2} \times \frac{5^{2}}{18^{2}}]

= 685660.1235 J = 685.66 k J

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