Answer:
Approximately [tex]36\; {\rm J}[/tex].
Explanation:
Under the assumption, there is no friction to hinder the motion of the block. As a result, the increase in the kinetic energy of the block would be equal to the external work done on the block.
Let [tex]F[/tex] denote this magnitude of this force, let [tex]s[/tex] denote the magnitude displacement, and let [tex]\theta[/tex] denote the angle between this force and displacement. To find the work that this force has done on the block, use the formula:
[tex]\begin{aligned}(\text{work}) &= F\, s\, \cos(\theta)\end{aligned}[/tex].
For example, since the block in this question is moving along a horizontal table, displacement of the block would be in the horizontal direction. It is given that the angle between this force and the horizontal direction is [tex]37^{\circ}[/tex]. Thus, the angle between the force and the displacement would be [tex]\theta = 37^{\circ}[/tex].
Magnitude of this force is [tex]F = 15\; {\rm N}[/tex], while magnitude of displacement is [tex]s = 3\; {\rm m}[/tex]. The work done on this block would be:
[tex]\begin{aligned}(\text{work}) &= F\, s\, \cos(\theta) \\ &= (15\; {\rm N})\, (3\; {\rm m})\, \cos(37^{\circ}) \\ &\approx 36\; {\rm N\cdot m} = 36\; {\rm J}\end{aligned}[/tex].
Hence, the kinetic energy of this block would increase by [tex]36\; {\rm J}[/tex].
