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? (b) What average force must he exert on the railing?

Answer 1

Answer

given,

mass of ice hockey player = 110 Kg

initial speed of the skate = 3 m/s

final speed of the skate = 0 m/s

distance of the center of mass, m = 30 cm = 0.3 m

a) Change in kinetic energy

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

    $\Delta KE = \dfrac{1}{2}m(0)^2 - \dfrac{1}{2}\times 110 \times 3^2$

    $\Delta KE = - 495\ J$

b) Average force must he exerted on the railing

     using work energy theorem

      W = Δ KE

      F .d  = -495

      F x 0.3  = -495

      F = -1650 N

the average force exerted on the railing is equal to 1650 N.

Answer 2

Answer:

(a). The change in the kinetic energy of his center of mass during this process is -495 J.

(b). The average force is 1650 N.

Explanation:

Given that,

Mass = 110 kg

Speed = 3.0 m/s

Distance = 30 cm

(a). We need to calculate the change in the kinetic energy of his center of mass during this process

Using formula of kinetic energy

$\Delta K.E=K.E_{2}-K.E_{1}$

$\Delta K.E=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2$

Put the value into the formula

$\Delta K.E=\dfrac{1}{2}\times110\times0^2-\dfrac{1}{2}\times110\times(3.0)^2$

$\Delta K.E=-495\ J$

(b). We need to calculate the average force must he exert on the railing

Using work energy theorem

$W=\Delta K.E$

$Fd=\Delta K.E$

$F=\dfrac{\Delta K.E}{d}$

Put the value into the formula

$F=\dfrac{-495}{30\times10^{-2}}$

$F=-1650\ N$

The average force is 1650 N.

Hence, (a). The change in the kinetic energy of his center of mass during this process is -495 J.

(b). The average force is 1650 N.