Question

A(n) 1610 kg car is moving along a level road at 21.7 m/s. The driver accelerates, and in the next 10 s the engine provides 22100 J of additional energy. If 3683.33 J of this energy must be used to overcome friction, what is the final speed of the car? Answer in units of m/s.

Answer 1

Answer:

v=22.22 m/s

Explanation:

Given that

m= 1610 kg

initial velocity ,u= 21.7 m/s

The energy provided by engine(W₁) = 22100 J

The energy lost in the friction(W₂) = 3683.33 J

Lets take final velocity of the car = v m/s

Now from energy conservation

Work done by all forces = Change in the kinetic energy

W₁ + W₂ = ΔKE

$W_1 - W_2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

By putting the values

$W_1 - W_2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$22100 - 3683.33=\dfrac{1}{2}\times 1610 v^2-\dfrac{1}{2}\times 1610 \times 21.7^2$

$\dfrac{1}{2}\times 1610 v^2=22100 - 3683.33+\dfrac{1}{2}\times 1610 \times 21.7^2$

v=22.22 m/s

This is final speed of the car.