Question

A 110 kg ice hockey player skates at 3.0 m/s toward a railing at the edge of the ice and then stops himself by grasping the railing with his outstretched arms. During the stopping process, his center of mass moves 30 cm toward the railing. (a) What is the change in the kinetic energy of his center of mass during this process

Answer 1

Answer:

Δ KE = -495 J

Explanation:

given,

mass of the ice hockey player = 110 Kg

initial speed = 3 m/s

final speed  = 0 m/s

distance, d = 0.3 m

change in kinetic energy

$\Delta K E = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$

$\Delta K E = \dfrac{1}{2}mv(0)^2 - \dfrac{1}{2}\times 110\times 3^2$

    Δ KE = -495 J

Hence, the change in kinetic energy is equal to Δ KE = -495 J