Answer to Question #97527 in General Chemistry for Keneen Maisonneuve

Question

A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it. (a) What happens to the potential energy of the ball as it is raised? (b) What quantity of work, in J, is used to raise the ball? (c) After the ball is dropped, it gains kinetic energy. If all the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the ball’s speed just before it hits the ground? (Note: The force due to gravity is F = m * g, where m is the mass of the object and g is the gravitational constant; g = 9.8 m>s2.

Answer 1

Mass of the ball = 5.4 Kg

Height = 1.6 m

(a)Let the potential energy of the ball at h=0 is 0.

The potential energy of the ball at h=1.6 is $-mgh=-(5.4 ×1.6×9.8) = - 84.672 J$ (Ans)

(b) $Work done = change in potential energy$

$= P.Ef - P.E.i$

$=(-84.672-0) J$

$= - 84.672 J$ (Ans)

(c)According to conservation of energy, $W = change$ in Kinetic Energy.

$K.E.f - K.E.i = W$

$K.E.i = 0 J$

$K.E.f = 1/2 mv2$

$1/2 mv2 = 84.672 J$

$v2 = 2×84.672/5.4$

$v = √914.4576 m/s$

$v = 30.24 m/s$ (Ans)