Question

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?

Answer 1

Answer:

Explanation:

Given

mass of block is m

inclination of ramp is $\theta$

Initial kinetic energy $\frac{1}{2}mv_0^2$

Length of ramp is L

Block move a distance of L so it moves a vertical distance of $L\sin \theta$

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

Initial kinetic energy $K.E._i=\frac{1}{2}mv_0^2$

Final kinetic energy $K.E._f=0$

$K.E._i-K.E._f=Work\ done\ by\ kinetic\ friction+work\ done\ by\ gravity$

$\frac{1}{2}mv_0^2=W_f+W_g$

$W_f=\frac{1}{2}mv_0^2-mgL\sin \theta$