A2.0 kg wood block is launched up a wooden ramp that is inclined at a 27 ∘ angle. the block's initial speed is 13 m/s . the coefficient of kinetic friction of wood on wood is μk=0.200. you may want to review (page 131) . for with math skills, you may want to review: vector components part a what vertical height does the block reach above its starting point? express your answer to two significant figures and include the appropriate units. view available hint(s) hh = nothing nothing part b what speed does it have when it slides back down to its starting point? express your answer to two significant figures and include the appropriate units. view available hint(s) vv = nothing nothing provide feedback

a)

m

b)

ms⁻¹

Explanation:

= mass of the wood block = 2 kg

= initial speed of the block at the bottom = 13 m/s

= angle of incline = 27°

= Coefficient of kinetic friction = 0.2

= kinetic frictional force

= height of the incline gained

= length of the incline traveled

In triangle ABC

= Normal force on the block by incline surface

From the force diagram of the block, force equation perpendicular to the incline surface is given as

                           Eq-1

kinetic frictional force on the block is given as

Using Eq-1

Work done by kinetic frictional force is given as

Using conservation of energy

Kinetic energy at the bottom = work done by frictional force + potential energy at the top

inserting the values

m

b)

= final speed of the block after returning to starting point

Using conservation of energy

Kinetic energy at the bottom initially = work done by frictional force + Kinetic energy at the bottom finally

inserting the values

ms⁻¹

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W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j