Question

Kramer goes bowling and decides to employ the force of gravity to "pick up a spare." He rolls the 7.0-kg bowling ball very slowly so that it comes to rest a center-to-center distance of 0.25 m from the one remaining 1.5-kg bowling pin.

Required:
Determine the force of gravity between the ball and the pin and comment on the efficacy of the technique.

Answer 1

Answer:

$1.12056\times 10^{-8}\ N$.

Kramer could try blowing air on the bowling pin.

Explanation:

$m_1$ = Mass of bowling ball = 7 kg

$m_2$ = Mass of bowling pin = 1.5 kg

$r$ = Distance between them = 0.25 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of gravity is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 7\times 1.5}{0.25^2}\\\Rightarrow F=1.12056\times 10^{-8}\ N$

The force of gravity between the ball and the pin is $1.12056\times 10^{-8}\ N$.

Kramer could try blowing air on the bowling pin.