Question

A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000 J of additional energy. If 3666.67 J of this energy must be used to overcome friction, what is the final speed of the car

Answer 1

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

• Mass of the car, m = 1700 kg
• Velocity of the car, v = 21 m/s
• Time of motion, t = 10 s
• Additional energy provided by the engine, E₁ = 22,000 J
• Energy used in overcoming friction, E₂ = 3,666.67 J

The change in the energy applied to the car is calculated as;

$\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta E = 18,333.33 \ J$

The final speed of the car is calculated as follows;

$\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2 = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s$

Thus, the final speed of the car at the given conditions is 30.1 m/s.

Learn more about change in kinetic energy here: brainly.com/question/6480366