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A very slippery block of ice slides down a smooth ramp tilted at angle ? The ice is released from rest at vertical height h above the bottom of the ramp.

A)Find an expression for the speed of the ice at the bottom?

B)Evaluate your answer to part a for ice released at a height of 22cm on ramp tilted at 40 ??

Respuesta :

Part 1

Here as we know by energy conservation

Initial potential energy + initial kinetic energy = final potential energy + final kinetic energy

now we will have

[tex]mgh_i + \frac{1}{2}mv_i^2 = mgh_f + \frac{1}{2}mv_f^2[/tex]

now we will plug in all data into the equation

so we will have

[tex]h_i = h[/tex]

[tex]v_i = 0[/tex]

[tex]h_f = 0[/tex]

from above equation now

[tex]mgh = \frac{1}{2}mv_f^2[/tex]

[tex]v_f = \sqrt{2gh}[/tex]

Part b)

as we know that

[tex]h = 22 cm[/tex]

[tex]\theta = 40 degree[/tex]

now from the above equation

[tex]v_f = \sqrt{2gh}[/tex]

[tex]v_f = \sqrt{2(9.81)(0.22)}[/tex]

[tex]v_f = 2.08 m/s[/tex]

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