Question

A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the instant at which his fingers lose contact with the wall, his center of mass has moved 0.50 m , and at this instant he is traveling at 2.5 m/s .

a)What is the average force exerted by the wall on him?
b)What is the work done by the wall on him?
c)What is the change in the kinetic energy of his center of mass?

Answer 1

(a). The average force exerted by the wall on him is 406.25 N.

(b). The work done by the wall on him is zero.

(c). The change in the kinetic energy of his center of mass is 203.1 J.

Step-by-step explanation:

Given that,

Mass of ice = 65 kg

Distance = 0.50 m

Speed = 2.5 m/s

(a). We need to calculate the average force exerted by the wall on him

Using work energy theorem

`W=\Delta K`

`Fd=(1)/(2)mv^2`

`F=((1)/(2)mv^2)/(d)`

`F=(mv^2)/(2d)`

Put the value into the formula

`F=(65*(2.5)^2)/(2*0.50)`

`F=406.25\ N`

(b). The wall does not move so the displacement is zero.

We need to calculate the work done by the wall on him

Using formula of work done

`W=Fd`

Put the value into the formula

`W=406.25*0`

`W=0`

(c). We need to calculate the change in the kinetic energy of his center of mass

Using formula of change in kinetic energy

`\Delta K=(1)/(2)mv^2`

Put the value into the formula

`\Delta K=(1)/(2)*65*(2.5)^2`

`\Delta K=203.1\ J`

Hence, (a). The average force exerted by the wall on him is 406.25 N.

(b). The work done by the wall on him is zero.

(c). The change in the kinetic energy of his center of mass is 203.1 J.