Answer :
Answer:
Approximately [tex]28.3\; \rm kg[/tex], assuming that the spring was pulled horizontally.
Explanation:
Let [tex]k[/tex] denote the spring constant of this spring. Calculate the size of tension in this spring that corresponds to the [tex]\Delta x = 0.88\; \rm m[/tex] stretch:
[tex]F(\text{tension}) = k\cdot \Delta x = 45\; \rm N \cdot m^{-1} \times 0.88\; \rm m = 39.6\; \rm N[/tex].
(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: [tex]39.6\; \rm N[/tex].
Newton's Second Law of motion suggests the following relation between the net force [tex]\sum F[/tex] on the object, the mass [tex]m[/tex] of the object, and the acceleration [tex]a[/tex] of the object:
[tex]\sum F = m \cdot a[/tex].
For this object, [tex]a = 1.4\; \rm m \cdot s^{-2}[/tex]. It has been found that the net force on this object is [tex]\sum F = 39.6\; \rm N[/tex]. Rearrange the original equation [tex]\sum F = m \cdot a[/tex] and solve for [tex]m[/tex], the mass of this object:
[tex]\displaystyle m = \frac{\sum F}{a} = \frac{39.6\; \rm N}{1.4\; \rm m \cdot s^{-2}} \approx 28.3\; \rm kg[/tex].
In other words (based on all these assumptions,) the mass of this object should be approximately [tex]28.3\; \rm kg[/tex].