Question

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the floor (Fig. P7.51). Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. Ignore friction and the mass of the pulley.

Answer 1

Explanation:

The given data is as follows.

    Mass of small bucket (m) = 4 kg

    Mass of big bucket (M) = 12 kg

    Initial velocity ($v_{o}$) = 0 m/s

    Final velocity ($v_{f}$) = ?

  Height $H_{o} = h_{f}$ = 2 m

and,    $H_{f} = h_{o}$ = 0 m

Now, according to the law of conservation of energy

         starting conditions = final conditions

  $\frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}$

     $\frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)$

                 235.44 = $8V^{2}_{f}$ + 78.48

                $V_{f}$ = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.