An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is 45 N/m and the spring is stretched by 0.88 m when the object is accelerating at 1.4 m/s2, what is the mass of the object?

Approximately , assuming that the spring was pulled horizontally.

Explanation:

Let denote the spring constant of this spring. Calculate the size of tension in this spring that corresponds to the stretch:

.

(The negative sign is ignored.)

If the spring is pulled horizontally, then the tension that the spring exerts would be the only unbalanced force on this object. The net force n this object would be equal to the tension that the spring had exerted: .

Newton's Second Law of motion suggests the following relation between the net force on the object, the mass of the object, and the acceleration of the object:

.

For this object, . It has been found that the net force on this object is . Rearrange the original equation and solve for , the mass of this object:

.

In other words (based on all these assumptions,) the mass of this object should be approximately .

answer: option c.

explanation:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j