Ablock of unknown mass is attached to a spring with a spring constant of 7.00 n/m 2 and undergoes simple harmonic motion with an amplitude of 12.5 cm. when the block is halfway between its equilibrium position and the end point, its speed is measured to be 32.0 cm/s (a )calculate the mass of the block.
( b)calculate the period of the motion.
(c ) calculate the maximum acceleration of the block.

Answers:

a)

b)

c)

Explanation:

We have the following data:

is the spring constant

is the amplitude of oscillation

is the velocity of the block when

Now let's begin with the answers:

We can solve this by the conservation of energy principle:

(1)

Where:

is the initial potential energy

 is the initial kinetic energy

is the final potential energy

is the final kinetic energy

Then:

(2)

Isolating :

(3)

(4)

(5)

The period is given by:

(6)

Substituting (5) in (6):

(7)

(8)

The maximum acceleration is when the force is maximum , as well :

(9)

Being

Hence:

(10)

Finding :

(11)

(12)

Finally:

-310J

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       (i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

In this case, the system is the balloon and;

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

Substitute these values into equation (i) as follows;

ΔE = -695 + 385

ΔE = -310J

Therefore, the change in internal energy is -310J

PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.