A sled and rider with a combined mass of 70 kg are at the top of a hill of height 9 m above the level ground below. The sled is given a push providing an initial kinetic energy of 1200 J at the top of the hill. What will be the kinetic energy of the sled and rider at the bottom of the hill, if friction can be ignored?

Respuesta :

Answer:

[tex]K_p=6174J[/tex]

Explanation:

Given:

[tex]m=70kg[/tex],[tex]m=9m[/tex],[tex]K_E=70kg[/tex]

[tex]K_E=\frac{1}{2}*m*v^2[/tex]

[tex]1200J=\frac{1}{2}*70kg*v^2[/tex]

[tex]v=5.8m/s[/tex]

To determine the kinetic energy at the bottom is:

[tex]K_p=m*g*h[/tex]

[tex]K_p=70kg*9.8m/s^2*9m[/tex]

[tex]K_p=6174J[/tex]

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