Answer
(a) $3.8m/s$
(b) $2.7m/s$
(c) $3.0m/s$
(d) points A and E
Work Step by Step
(a)We can find the object's speed at point B as follows:
$K.E_i+U_i=K.E_f+U_f$
$\implies 0+10J=\frac{1}{2}mv_f^2+U_f$
This simplifies to:
$v_f=\sqrt{\frac{2(10J-U_f)}{m}}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{2(10J-2.0J)}{1.1Kg}}$
$v_f=3.8m/s$
(b) The object's speed at point C can be calculated as:
$v_f=\sqrt{\frac{2(10J-U_f)}{m}}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{2(10J-6.0J)}{1.1Kg}}$
$v_f=2.7m/s$
(c) The object's speed at point D can be calculated as:
$v_f=\sqrt{\frac{2(10J-U_f)}{m}}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{2(10J-5.0J)}{1.1Kg}}$
$v_f=3.0m/s$
(d) We know that the object starts at rest at point A. This is the point where the object has the maximum potential energy and minimum kinetic energy. From this point onward when the object travels the energy changes, so this is the turning point. Similarly, point E is at the same level as point A. Thus, point E is the second turning point.
