Answer
a) $d=0.937m$
b) $W_c=16.5J$
c) $W_{nc}=-7.72J$
d)$\Delta K.E=8.8J$
$W_c=16.5J$
$\Delta E=-7.7J$
Work Step by Step
(a) We can find the required distance as follows:
$d=\frac{(m_1+m_2)v_f^2}{2g(m_2-\mu_k m_1)}$
We plug in the known values to obtain:
$d=\frac{(2.40+1.80)(2.05)}{2(9.81)[1.80-(0.350)(2.40)]}$
$\implies d=0.937m$
(b) We can find the conservative work done on this system as
$W_c=-m_2g\Delta y$
We plug in the known values to obtain:
$W_c=-(1.80)(9.81)(-0.937)$
$W_c=16.5J$
(c) We can find the non-conservative work done on this system as
$W_{nc}=-\mu_k m_1gd$
We plug in the known values to obtain:
$W_{nc}=-(0.350)(2.40)(9.81)(0.937)$
$W_{nc}=-7.72J$
(d) We know that
$\Delta K.E=\frac{1}{2}(m_1+m_2)(v_f^2-v_i^2)=\frac{1}{2}(2.40+1.80)[(2.05)^2-(0)^2]$
$\Delta K.E=8.83J$
Now $\Delta K.E=W_c+W_{nc}=16.5-7.72=8.8J$
where $W_c=16.5J=-\Delta U$
$\Delta E=\Delta K.E+\Delta U=8.83+(-16.5)=-7.7J$
