Suppose that 6 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 43 cm. (a) How much work is needed to stretch the spring from 33 cm to 38 cm

Workdone = 0.67 Joules

Explanation:

Given the following data;

Workdone = 6J

Extension = 43 - 28 = 15cm to meters = 15/100 = 0.15m

The workdone to stretch a string is given by the formula;

Workdone = ½ke²

Where;

k is the constant of elasticity.

e is the extension of the string.

We would solve for string constant, k;

6 = ½*k*0.15²

6 = ½*k*0.0225

Cross-multiplying, we have;

12 = 0.0225k

k = 12/0.0225

k = 533.33N/m

a. To find the workdone when e = 33, 38.

Extension = 38 - 33 = 5 to meters = 5/100 = 0.05m

Workdone = ½*533.33*0.05²

Workdone = 266.67 * 0.0025

Workdone = 0.67 Joules

Therefore, the amount of work (in J) needed to stretch the spring from 33 cm to 38 cm is 0.67.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.