contestada

In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills.) To demonstrate this, find the final speed in m/s and the time taken in seconds for a skier who skies 69.0 m along a 28° slope neglecting friction for the following cases.(a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events.

Respuesta :

Answer:

Part a)

[tex]v_f = 25.2 m/s[/tex]

[tex]t = 5.48 s[/tex]

Part b)

[tex]v_f = 25.32 m/s[/tex]

[tex]t = 4.96 s[/tex]

Explanation:

Part a)

When ski start from rest

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

on this inclined plane we know that the acceleration is given as

[tex]a = g sin\theta[/tex]

[tex]a = 9.81 sin28[/tex]

[tex]a = 4.6 m/s^2[/tex]

now for final speed

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 0 = 2(4.6)(69)[/tex]

[tex]v_f = 25.2 m/s[/tex]

now time taken by the ski to reach the bottom is given as

[tex]v_f = v_i + at[/tex]

[tex]25.2 = 0 + 4.6 t[/tex]

[tex]t = 5.48 s[/tex]

Part b)

Now when ski start with initial speed of 2.5 m/s

then we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 2.5^2 = 2(4.6)(69)[/tex]

[tex]v_f = 25.32 m/s[/tex]

now time taken by the ski to reach the bottom is given as

[tex]v_f = v_i + at[/tex]

[tex]25.32 = 2.5 + 4.6 t[/tex]

[tex]t = 4.96 s[/tex]

Otras preguntas