contestada

8. A 55-kg skateboarder enters a ramp moving horizontally with a speed of 6.5 m/s, and leaves the ramp moving vertically with a speed of 4.1 m/s. (A) Find the height of the ramp, assuming no energy loss to frictional forces. (B) What is the skateboarder's maximum height above the bottom of the ramp? ​

Respuesta :

onyebuchinnaji

(a) The energy lost to friction by the skateboarder is -699.6 J.

(b) The maximum height above the bottom of the ramp is 1.3 m.

Energy loss to friction

The energy lost to friction by the skateboarder is calculated from the change in kinetic energy of the skateboarder.

ΔK.E = K.Ef - K.Ei

ΔK.E = ¹/₂m(vi² - vf²)

ΔK.E = 0.5 x 55 (4.1² - 6.5²)

ΔK.E = -699.6 J

Maximum height above the bottom of the ramp

The maximum height above the bottom of the ramp is determined from conservation of energy.

mghi + ¹/₂mvi² = ¹/₂mvf² + mghf

55 x 9.8 x hi  +  (0.5 x 55 x 6.5²) = (0.5 x 55 x 4.1²) + (55 x 9.8 x 0)

539hi + 1161.88 = 462.28

539hi =  -699.6

hi = -699.6/539

hi = -1.3 m

Thus, the maximum height above the bottom of the ramp is 1.3 m.

Learn more about conservation of energy here: https://brainly.com/question/166559

Otras preguntas