Question

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of vi=2.62 m/s. To reach the rack, the ball rolls up a ramp that rises through a vertical distance of h=0.47m. What is the linear speed of the ball when it reaches the top of the ramp?

Answer 1

Answer:

vf = 4.01 m/s

Explanation:

According to the law of conservation of energy:

$Kinetic\ Energy\ Lost\ by\ Ball = Potential\ Energy\ Gained\ by\ the\ Ball \\\frac{1}{2}m(v_f^2-v_i^2) = mgh\\\\ v_f^2-v_i^2 = 2gh\\v_f^2 = 2gh + v_i^2$

where,

vf = final speed of ball at top of the ramp = ?

vi = initial speed of ball = 2.62 m/s

g = acceleration due to gravity = 9.81 m/s²

h = height = 0.47 m

Therefore,

$v_f^2 = (2)(9.81\ m/s^2)(0.47\ m)+(2.62\ m/s)^2\\v_f = \sqrt{9.2214\ m^2/s^2+6.8644\ m^2/s^2}$

vf = 4.01 m/s