Question

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?

Answer 1

Approximately
`28.3\; \rm kg`, assuming that the spring was pulled horizontally.

Step-by-step explanation:

Let
`k` denote the spring constant of this spring. Calculate the size of tension in this spring that corresponds to the
`\Delta x = 0.88\; \rm m` stretch:

`F(\text{tension}) = k\cdot \Delta x = 45\; \rm N \cdot m^(-1) * 0.88\; \rm m = 39.6\; \rm N`.

(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:
`39.6\; \rm N`.

Newton's Second Law of motion suggests the following relation between the net force
`\sum F` on the object, the mass
`m` of the object, and the acceleration
`a` of the object:

`\sum F = m \cdot a`.

For this object,
`a = 1.4\; \rm m \cdot s^(-2)`. It has been found that the net force on this object is
`\sum F = 39.6\; \rm N`. Rearrange the original equation
`\sum F = m \cdot a` and solve for
`m`, the mass of this object:

`\displaystyle m = (\sum F)/(a) = (39.6\; \rm N)/(1.4\; \rm m \cdot s^(-2)) \approx 28.3\; \rm kg`.

In other words (based on all these assumptions,) the mass of this object should be approximately
`28.3\; \rm kg`.