Question

5. A roller coaster extends to the groundfrom a height of 39 m (point A) and then risesto a height of 11 m (point B). An object of mass 12 kgstarts at point A with a speed of 7 m/s. Assumingthe roller coaster is friction less, calculate thespeed of the object by the time it reaches point B. (1 point)A. O 14.231 m/sB. O 29.606 m/sC. O 32.44 m/sD. O 24.45 m/sE. O 3.293 m/s

Answer 1

Apply:

potential energy A + kinetic energy A = potential energy B + kinetic energy B

m g ha + 1/2 m va^2 = m g hb + 1/2 m vb^2

Where:

m= 12kg

g= 9.8 m/s^2

ha = height A = 39m

va = speed A = 7m/s

hb = height B = 11 m

vb = speed B = ?

Replacing with the values given:

(12)(9.8)(39) + 1/2 (12) (7)^2 = (12)(9.8)(11) + (1/2)(12)(vb)^2

4,586.4 + 294 = 1,293.6 + 6 vb^2

4,880.4 - 1,293.6 = 6vb^2

3586.8 = 6vb ^2

3586.8/6 = vb^2

597.8 = vb^2

√597.8 = vb

vb = 24.45 m/s (option D)