Two balls, one of gold with a mass of 100 kg and one of wood with a mass of 1 kg, are suspended 1 meter apart. What is the attractive force, in newtons, of
a. the gold ball acting on the wooden ball?
b. the wooden ball acting on the gold ball?
a,b both forces are equal and opposite.
a. force=GMm/distance^2
I solve it, but i don't know the sign of each one of them .. can you help me in this part please ..
F = (G.m1.m2) / (r2)
F = (6.673 x 10-11 x 100 x 1) / (12)
F = 6.673 x 10-9 N
To calculate the attractive force between the two balls, we can use Newton's law of universal gravitation, which states that the force of gravitational attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Let's calculate the attractive force for each case:
a. The Gold ball acting on the Wooden ball:
The mass of the Gold ball is 100 kg, and the mass of the Wooden ball is 1 kg. The distance between them is 1 meter.
Using the formula:
F = (G * m1 * m2) / d^2
Where:
F is the attractive force
G is the gravitational constant (approximately 6.674 x 10^-11 N(m/kg)^2)
m1 and m2 are the masses of the objects
d is the distance between the centers of the objects
Plugging in the values:
F = (6.674 x 10^-11 N(m/kg)^2 * 100 kg * 1 kg) / (1 meter)^2
F = 6.674 x 10^-11 N(m/kg)^2 * 100 kg
F = 6.674 x 10^-9 N
So, the attractive force of the Gold ball acting on the Wooden ball is approximately 6.674 x 10^-9 N.
b. The Wooden ball acting on the Gold ball:
To calculate the force exerted by the Wooden ball on the Gold ball, we apply Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Since the force exerted by the Gold ball on the Wooden ball was calculated as 6.674 x 10^-9 N, the force exerted by the Wooden ball on the Gold ball will be the same.
Therefore, the attractive force of the Wooden ball acting on the Gold ball is approximately 6.674 x 10^-9 N as well.