Jay rides his 2.0-kgkg skateboard. He is moving at speed 5.8 m/sm/s when he pushes off the board and continues to move forward in the air at 5.4 m/sm/s. The board now goes forward at 13 m/sm/s. Part APart complete Determine Jay's mass. Express your answer to two significant figures and include the appropriate units. mJmJ = 36 kgkg SubmitPrevious Answers Correct Part B Determine the change in the internal energy of the system during this process.

36 kg, - 54.72 J

Explanation:

Part A.

From the general momentum equation we have that:

Pji + Psi + Jonj + Jons = Pjf + Psf

Let's just consider the x axis:

Pjix + Psix + Jx = Pjfx + Psfx

Pjix = Mj * vix

Psix = Ms * vix

Mj = what we want to know

Ms = 2 kg

vix = 5.8 m / s

Jx = 0

Pjfx = Mj * vjfx

Psfx = Ms * vsfx

vsfx = 13 m / s

vjfx = 5.4 m / s

By replacing all these values we are left with:

Mj * 5.8 + 2 * 5.8 + 0 = 5.4 * Mj + 2 * 13

0.4 * Mj = 26 - 11.6

Mj = 14.4 / 0.4

Mj = 36 kg

Part B.

If we apply the generelized work energy principle we have to:

Ui + W = Uf

Ki + 0 = Kf + Uint

Ki = 1/2 * (Mj + Ms) * vix ^ 2

All known values, if we replace we have:

Ki = (1/2) * (36 + 2) * (5.8 ^ 2)

Ki = 639.16

Kf = 1/2 * (Mj * vjfx ^ 2 + Ms * vsfx ^ 2)

Replacing the values, we are left with:

Kf = 1/2 * (36 * (5.4) ^ 2 + 2 * (13 ^ 2))

Kf = 693.88

Replacing in:

Ki + 0 = Kf + Uint

Uint = Ki - Kf

Uint = 639.16 - 693.88

Uint = - 54.72

The internal energy is - 54.72 Jules

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