Answer:
Approximately [tex](-6)\; {\rm J}[/tex], assuming that [tex]g = 9.81\; {\rm m \cdot s^{-2}}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this box.
Decompose the weight [tex]m\, g[/tex] of the box into two components: perpendicular to the slope [tex](-m\, g)\, \cos(15^{\circ})[/tex] and parallel to the slope [tex](-m\, g)\, \sin(15^{\circ})[/tex]. The perpendicular component is balanced with the normal force of the slope on the box. In contrast, since there is no friction between the box and the slope, the parallel component isn't balanced and will be equal to the net force on the box:
[tex]F_{\text{net}} = (-m\, g) \sin(15^{\circ}})[/tex].
Note that [tex]F_{\text{net}}[/tex] is constant during this [tex]t = 1\; {\rm s}[/tex].
The acceleration [tex]a[/tex] of this box will be:
[tex]\begin{aligned}a &= \frac{F_{\text{net}}}{m}\\ &= \frac{(-m\, g) \sin(15^{\circ})}{m} \\ &= (-g)\, \sin(15^{\circ})\end{aligned}[/tex].
Let [tex]u[/tex] denote the initial velocity of this box. Since after the [tex]t = 1\; {\rm s}[/tex] of acceleration at [tex]a = (-g)\, \sin(15^{\circ}})[/tex] the velocity of the box is [tex]v = 0\; {\rm m\cdot s^{-1}}[/tex], the initial velocity of this box will be:
[tex]\begin{aligned} u &= v - a\, t \\ &= 0 - (-g)\, \sin(15^{\circ})\, t \\ &= g\, t\, \sin(15^{\circ}})\end{aligned}[/tex].
Initial kinetic energy of this box would be:
[tex]\begin{aligned} (\text{KE}) &= \frac{1}{2}\, m\, u^{2} \\ &= \frac{1}{2}\, m\, (g\, t\, \sin(15)^{\circ}})^{2}\end{aligned}[/tex].
The kinetic energy of this box is [tex]0[/tex] when the box is at rest. The change in kinetic energy was [tex](1/2)\, m\, (g\, t\, \sin(16^{\circ}))^{2}[/tex]. That energy change should be the opposite of the work the external force weight did on this box.
Substitute in [tex]m = 2\; {\rm kg}[/tex], [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex], [tex]t = 1\; {\rm m\cdot s^{-1}}[/tex] to find the change in kinetic energy and hence the work that the net force did on the box:
[tex]\begin{aligned} (\text{Work Done}) &= -(\text{Kinetic energy change}) \\ &= -\frac{1}{2}\, m\, (g\, t\, \sin(15)^{\circ}})^{2} \\ &= -\frac{1}{2}\times 2\; {\rm kg} \times (9.81\; {\rm m\cdot s^{-2}} \times 1\; {\rm s} \times \sin(15^{\circ})^{\circ} \\ &\approx (-6)\; {\rm J}\end{aligned}[/tex].
