Question

# A block of density $$2000 kg/m^{3}$$ and mass $$10 kg$$ is suspended by a  spring of stiffness $$100 N/m$$. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density $$1000 kg/m^{3}$$ . If the block is in equilibrium position.

A
the elongation of the spring is 1cm.
B
the magnitude of buoyant force acting on the block is 50N.
C
the spring potential energy is 12.5J.
D
magnitude of spring force on the block is greater than the weight of the block.

Solution

## The correct options are A the magnitude of buoyant force acting on the block is $$50 N.$$ B the spring potential energy is $$12.5 J.$$There are three forces when the block is in equilibrium.1. weight of the block downward.2. buoyancy force due to the liquid upward3. force due to the spring upwardWeight of block: $$m_bg=V_b\rho_bg=V_b\times 2000\times 10$$Buoyancy force upward: $$F_{buoyancy}=V_b\rho_lg=V_b\times 1000\times 10$$Upward force due to spring: $$F_{spring}=kx=100x$$Equating the upward force with the downward force,$$V_b\times 1000\times 10 + 100x =V_b\times 2000\times 10$$$$\therefore 100x = V_b \times 10000= \dfrac{10}{2000} \times 10000$$$$\therefore x= \dfrac{1}{2}\:m$$Potential Energy stored in the spring: $$PE = \dfrac{1}{2}kx^2= \dfrac{1}{2} \times 100 \times ( \dfrac{1}{2})^2= 12.5\: Nm = 12.5 \:J$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More