Question

A wooden ball with a weight of 18.0 N hangs from a string tied to a spring scale. When the ball is at rest, exactly 50% submerged in water, the spring scale reads 5.00 N. For this problem, we will use a density of water of 1000 kg/m3, and we will use g = 10.0 m/s2. (a) If the ball was only 20.0% submerged in water instead, what would the spring scale read? (b) Determine the density of the ball. (c) Determine the volume of the ball.

Answer 1

a)
`12.8N`

b)
`692.308(kg)/(m^(3))`

c)
`2.6(10^(-3))m^(3)`

Step-by-step explanation:

Let's write the information given by the exercise.

`W(WoodenBall)=18.0N`

δ (Water) =
`1000(kg)/(m^(3))`

`g=10.0(m)/(s^(2))`

The buoyant force exerted by the water is :

E= V.δ.g

Where E is the buoyant force

V is the shifted volume of the fluid

δ is the fluid density

and g is the gravity acceleration.

We can write the following equation :

`W(WoodenBall)-E=5.00N`

18.0 N - V.δ(water).g = 5.00 N

`13N=0.5[V(WoodenBall)]1000(kg)/(m^(3)).10(m)/(s^(2))`

Let's remember that

`1N=1(kg.m)/(s^(2))`

Therefore,

`13N=0.5[V(WoodenBall)].10000(N)/(m^(3))`

`V(WoodenBall)=2.6(10^(-3))m^(3)`

And that is the answer for c)

For a) we write

`SpringScaleReading=W(WoodenBall)-E`

Spring scale reading = 18 N - 0.2 [V(Wooden ball)] . δ(water). g

`SpringScaleReading=18N-0.2[2.6(10^(-3))m^(3)].1000(kg)/(m^(3)).10(m)/(s^(2))`

`SpringScaleReading=18N-5.2N=12.8N`

For a) The spring scale will read 12.8 N

b)
`W(WoodenBall)=m(WoodenBall).g`

Where m(Wooden Ball) is the mass of the wooden ball.

`18N=m(WoodenBall).10(m)/(s^(2))`

`m(WoodenBall)=1.8kg`

δ(Wooden ball) = m(Wooden ball) / V(Wooden ball)

δ(Wooden ball) = 1.8 kg /
`2.6(10^(-3))m^(3)`

δ(Wooden ball) =
`692.308 (kg)/(m^(3))`