Respuesta :
As we know that the work done is given as
[tex]W = Fdcos\theta[/tex]
here we know that
W = 5600 J
F = 150 N
[tex]\theta = 45 degree[/tex]
now we will have
[tex]5600 = 150 \times d \times cos45[/tex]
Now solving for distance "d"
[tex]d = \frac{5600}{150\times cos45}[/tex]
[tex]d = 5.3 \times 10^1 m[/tex]
so above is the horizontal distance moved