Question

You drop a 2.00 kg book to a friend who stands on the ground atdistance D = 10.0 m below. Your friend's outstretchedhands are at distance d =1.20 m above the ground (Fig.8-29). (a) What is the speed of the book when itreaches her hands? (b) If we substituted asecondbook with twice the mass, what would its speed be when it reachesher hands? (c) If, instead, the book were throwndown with initial speed5.70 m/s, what would be the answer to(a)?

Answer 1

Note: For this problem, we will use conservation of energy

(a)





Note: The mass of the book does not matter when using energyconservation, thus, it cancels out
Note: The initial kinetic energy is zero because it was dropped







(b)

As stated previously in part (a), the mass of the book does notmatter, thus, its speed would be the same as it was in part (a)

(c)

Answer 2

(a)

s = 10meters
v(initial) = 0m/s
v(final) = ?
a = 9.8m/s² (assume down is positive)


v(final)² = v(initial)² +2as
v(final)² = 0² +2(9.8)(10)
v(final) = 14 m/s

(b)

same v(final) as in part (a) as changing mass of an object does notchange velocity

proof:

F = mass x acceleration (Newton'ssecond law)

force on an object falling is F = mass xgravity

so you can write:

F = ma
mass x gravity = mass x acceleration
now mass will cancel out from both side

gravity = acceleration

hence acceleration (change is velocity) does not depend on mass

(c)
s = 10meters
v(initial) = 5.70m/s
v(final) = ?
a = 9.8m/s² (assume down is positive)

v(final)² = v(initial)² +2as
v(final)² = 5.70² +2(9.8)(10)
v(final) = 15.1158857m/s