Problem 6.29

6.29. a)

We know that the work done by gravity is given by

$W_{g} = mg\ell \cos (90{}^{\circ}+ 20{}^{\circ})$

$W_{g} = (10 kg)(9.8 m/s^{2})(5 m)(\cos 110{}^{\circ}) ~=~ -168 J$

b)

The work done by friction is given by

$W_{f} = -\ell f_{k}$

$W_{f} = - \ell \mu_{k} mg\cos \theta$

$W_{f} = - (5 m)(0.4)(10)(9.8)\cos 20{}^{\circ}= -184 J$

c)

The work done by the 100 N force is given by

$W_{F} = F \ell ~=~ (100 N)(5 m) ~=~ 500 J$

d)

The change in kinetic energy is given by

$\Delta K = W_{net} = W_{F} + W_{f} + W_{g}$

$\Delta K = 148 J$

e)

We know that

$\Delta K = K_{f} - K_{i} ~=~ \frac{1}{2} m (v_{f}^{2} - v_{0}^{2})$

or solving for v_f we find that

$v_{f} = \sqrt{\frac{2(\Delta K)}{m} + v_{0}^{2}}$

$v_{f} = \sqrt{\frac{2(148)}{10} + (1.5)^{2}} ~=~ 5.64 m/s$