# Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises - Page 853: 50

$\approx 1299038 J$

#### Work Step by Step

Horizontal component of the force is: $F=1500 cos 30 ^\circ$ $F=1500 cos 30 ^\circ=1500 \times \frac{\sqrt 3}{2}=750 \sqrt 3 N$ $d =1 km =1000 m$ As we know the work done is calculated as: $W= F.d$ Thus, $W=( 750 \sqrt 3 N) (1000 m)$ $W= 750000\sqrt 3 Nm$ $W \approx 1299038 J$

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