Problem 31P

Archimedes’ principle can be used not only to determine the specific gravity of a solid using a known liquid (Example 10–8); the reverse can be done as well. (a) As an example, a 3.40-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the density of the liquid. (b) Derive a formula for determining the density of a liquid using this procedure.

EXAMPLE 10–8

FIGURE 10–14 (a) A scale reads the mass of an object in air—in this case the crown of Example 10–8. All objects are at rest, so the tension FT in the connecting cord equals the weight w of the object: FT = mg. We show the free-body diagram of the crown, and FT is what causes the scale reading (it’s equal to the net downward force on the scale, by Newton’s third law). (b) Submerged, the object has an additional force on it, the buoyant force FB. The net force is zero, so F′T + FB = mg (= w). The scale now reads m′ = 13.4 kg, where m′ is related to the effective weight by w′ = m′g. Thus F′T = w′ = w – FB.

TABLE 10–1 Densities of Substances†

ANSWER:STEP 1:-The spring constant of the spring is, K = 250 N/m.The mass of the cart is 10 kg.The displacement of the spring is 60 cm = 0.6 m.We know that, every ideal spring obeys Hooke’s law and the potential energy of the springaccording to the law is, P.E = 1/2 Kx = 1/2 × 250 × 0.6 = 45jules.STEP 2:-Now let’s use the law of mechanical energy conservation, K.E + P.E = K.E + P.E -------------------------(1) initial initial final final 2 2 2The initial kinetic energy is, 1/2 mv = 1/2 × 10 × v = 5 × v , where v is the initial velocity.Initial potential energy is zero, as the displacement is zero.Final kinetic energy is zero, as the cart comes to rest at the end. 2Final potential energy is P.E = 1/2 Kx = 1/2 × 250 × 0.6 = 45julesNow let’s put the values in equation (1), K.E + P.E = K.E + P.E initial initial final final 5 × v + 0 = 0 + 45 5 × v = 45\n v = 45 / 5 = 9 v = 9 = 3 m/s.CONCLUSION:So, the cart must have come with a speed of 3 m/s towards the spring.