A water-skier of mass 75.0 kg initially at rest is being pulled due east by a horizontal towrope. The rope exerts a force of 365 N (east). The water (and air) exerts a combined average frictional force of 190 N (in the opposite direction). How fast will the skier be moving after a distance of 38.0 m?

Respuesta :

Answer:

The skier will be moving at 13.31 m/s.

Explanation:

To calculate the velocity of the skier we need to find the acceleration, as follows:

[tex] \Sigma F = ma [/tex]

[tex] F_{r} - F_{f} = ma [/tex]

Where:

[tex] F_{r}[/tex]: is the force due to the rope = 365 N

[tex] F_{f}[/tex]: is the combined average frictional force = 190 N

m: is the mass = 75.0 kg

[tex] a = \frac{365 N - 190 N}{75.0 kg} = 2.33 m/s^{2} [/tex]

Now, we can calculate the velocity of the skier by using the following kinematic equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex] v_{f}[/tex]: is the final velocity =?

[tex] v_{0}[/tex]: is the initial velocity = 0 (the skier is initially at rest)

d: is the distance = 38.0 m

[tex] v_{f} = \sqrt{2*2.33 m/s^{2}*38.0 m} = 13.31 m/s [/tex]

Therefore, the skier will be moving at 13.31 m/s.

I hope it helps you!