What is the magnitude of the initial acceleration of the sphere when it is released?

Question

A small sphere of wood with a density p=0.40 g/cm^3 is held at rest well under the surface of a pool of water. What is the magnitude of the initial acceleration of the sphere when released?

Possible Ans:
a.)15 m/s^2
b.)9.8 m/s^2
c.)33 m/s^2
d.)23 m/s^2
e.)3.4 m/s^2

Answer 1

Effectively, putting the sphere under the water is the same as exchanging a sphere at the surface for an equivalent volume of water at the given depth. The sphere loses gravitational PE (= mgh, where h is the depth) and the water gains it; since the sphere is lighter than the water a net input of energy is required.

To find the force on the sphere, remember that force is the derivative of potential energy with respect to position; if this is a bit too scary, think of E = Fd for constant force. Here we have E = (m_water - m_wood).g.h, so the net force is (m_water - m_wood).g.

Now, we know m_water = m_wood * 2.5 (i.e. 1.0 / 0.40), and the acceleration is given by F = ma, so we have
F = m_wood * (2.5 - 1.0) * g = m_wood * a
and so a = 1.5 * g = 14.7 m/s^2 or 15 m/s^2 to 2 s.f.