(a) In Problem 3, what is the speed of the book when it reaches the hands? (b) If we substituted a second book with twice the mass, what would its speed be? (c) If, instead, the book were thrown down, would the answer to (a) increase, decrease, or remain the same?

i) Mass of book_{$m=2\text{\hspace{0.17em}kg}$}

ii) Height from the ground _{$\left(D\right)=10\text{m}$}

iii) Stretched hand at distance _{$\left(d\right)=1.50\text{m}$}

iv) Gravitational acceleration $\left(D\right)=9.80{\text{m/s}}^{\text{2}}$

The mechanical energy _{$E_{mec}$} of a system is the sum of its kinetic energyand potential energy.

Formula:

i)

ii)

iii)

$h=D-d$

$h=10-1.50$

$h=8.50\text{m}$

From problem 3

_{$U_i=mgD$}

_{$\Rightarrow U_i=2\times 9.80\times 10$}

_{$\Rightarrow U_i=196\text{J}$}

_{$\Rightarrow U_f=mgd$}

_{$\Rightarrow U_f=2\times 9.80\times 1.5$}

_{$\Rightarrow U_f=29\text{\hspace{0.17em}J}$}

By using law of conservation of mechanical energy,

_{$K_i+U_i=K_f+U_f$}

_{$\Rightarrow 0+196=K_f+29$}

_{$\begin{array}{c}{K}_f=196-29\\ =167\text{\hspace{0.17em}J}\end{array}$}

$\Rightarrow K_f=\frac{1}{2}m{v}_f^2$

_{$\Rightarrow v_f=\sqrt{\frac{2K_f}{m}}$}

_{$\Rightarrow v_f=\sqrt{\frac{2\times 167}{2}}$}

_{$\Rightarrow v_f=\sqrt{167}$}

_{$\Rightarrow v_f=12.9\text{m/s}$}

If mass is doubled, then$K_f$is also doubled.

$K_f=2\times 167$

$K_f=m{v}_f^2$

$\Rightarrow v_f=\sqrt{\frac{2K_f}{2m}}$

$\Rightarrow v_f=\sqrt{\frac{2\times 2\times 167}{2\times 2}}$

$\Rightarrow v_f=\sqrt{167}$

$\Rightarrow v_f=12.9\text{m/s}$

If$K_i\ne 0,$we find$K_f=mgh+K_i$ . The value of$K_i$is always positive. This would result in a larger value for$K_f$than in the previous parts, and thus it leads to a larger value for_{$v_f$}.