Two identical objects are pressed against two different springs so that each spring stores 55.0J of potential energy. The objects are then released from rest. One spring is quite stiff (hard to compress), while the other one is quite flexible (easy to compress). Which of the following statements is or are true? (More than one statement may be true.) A. Both objects will have the same maximum speed after being released. B. The object pressed against the stiff spring will gain more kinetic energy than the other object. C. Both springs are initially compressed by the same amount. D. The stiff spring has a larger spring constant than the flexible spring. E. The flexible spring must have been compressed more than the stiff spring.

A, D, E

Explanation:

Solution:-

- Consider the two identical objects with mass (m).

- The stiffness of the springs are (k1 and k2).

- Both the spring store 55.0 J of potential energy.

- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.

- The maximum speed (v) is attained when all the potential energy is converted to kinetic energy.

- Apply Energy conservation for spring with stiffness (k1).

ΔU = ΔEk

55.0 = 0.5*m*v^2

v = √ (110 / m)

- Apply Energy conservation for spring with stiffness (k2).

ΔU = ΔEk

55.0 = 0.5*m*v^2

v = √ (110 / m)

Answer: Both objects will have the same maximum speed (A)

- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:

k1 > k2

Where,

k1: The stiff spring

k2: The flexible spring

Answer: The stiff spring has a larger spring constant than the flexible spring. (D)

- We will assume that the spring with constant (k1) undergoes a displacement (x1) and the spring with constant (k2) undergoes a displacement (x2). The potential energy stored in both spring is the same. Hence,

U1 = U2

0.5 * (k1) * (x1) ^2 = 0.5 * (k2) * (x2) ^2

[ k1 / k2 ] = [ x2 / x1 ]^2

Since,

k1 > k2, then [ k1 / k2 ] > 1

Then,

[ x2 / x1 ]^2 > 1

[ x2 / x1 ] > 1

x2 > x1

Answer: The flexible spring (x2) was compressed more than the stiff spring (x1). (E)