A block of mass m slids up a ramp making an angle θ with the horizontal. The block has an initial KE of 1 2 m v2 0 as it starts up the ramp and travels a distance L along the ramp before coming to momentary rest. v0 L How much work did kinetic friction do on the block between its starting point and the point it came to momentary rest?

Respuesta :

Answer:

Explanation:

Given

mass of block is m

inclination of ramp is [tex]\theta [/tex]

Initial kinetic energy [tex]\frac{1}{2}mv_0^2[/tex]

Length of ramp is L

Block move a distance of L so it moves a vertical distance of [tex]L\sin \theta[/tex]

Applying work Energy theorem i.e. change in kinetic energy of object is equal to work done by all the forces

Initial kinetic energy [tex]K.E._i=\frac{1}{2}mv_0^2 [/tex]

Final kinetic energy [tex]K.E._f=0[/tex]

[tex]K.E._i-K.E._f=Work\ done\ by\ kinetic\ friction+work\ done\ by\ gravity[/tex]

[tex]\frac{1}{2}mv_0^2=W_f+W_g[/tex]

[tex]W_f=\frac{1}{2}mv_0^2-mgL\sin \theta [/tex]

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