Question

A 2300 kg car moving at an initial speed of 25 m/s along a horizontal road skids to a stop in 50 m. (Note: When stopping without skidding and using conventional brakes, 100 percent of the kinetic energy is dissipated by friction within the brakes. With regenerative braking, such as that used in hybrid vehicles, only 70 percent of the kinetic energy is dissipated.)

(a) Find the energy dissipated by friction. 718750 Incorrect: Your answer is incorrect. kJ

(b) Find the coefficient of kinetic friction between the tires and the road.

A 0.17 kg baseball is launched from the roof of a building 11 m above the ground. Its initial velocity is 25 m/s at 40° above the horizontal. Assume any effects of air resistance are negligible.

(a) What is the maximum height above the ground that the ball
reaches?

(b) What is the speed of the ball as it strikes the ground?

Answer #1

u = 25 m/s

Distance = 50 m

Energy Dissipated by Friction = 70% * Kinetic Energy

Kinetic Energy = 0.5 * m*v^2

Kinetic Energy = 0.5 *2300*25^2

Kinetic Energy = 718750 J

Energy Dissipated by Friction = (70 *718750)/ 100

Energy Dissipated by Friction = 503125 J

**Energy Dissipated by Friction = 503.125 KJ**

(b)

Force * Displacement = Work done

Force = 503125 / 50

Force = 10062.5 N

Force = u*m*g

u*2300*9.8 = 10062.5

u = 0.446

**Coefficient of kinetic friction between the tires and the
road, u = 0.446**

*Please post seperate questions in seperate post.*

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