The work done by a force is given by:
![W=Fd \cos \theta](https://img.qammunity.org/2019/formulas/physics/high-school/ep8u09fspchztdhzo80usjdsjncp7yzsag.png)
where
F is the magnitude of the force
d is the distance covered by the object
![\theta](https://img.qammunity.org/2019/formulas/physics/high-school/4mdcq65qvll7syb3jnyuizz28syhprwuug.png)
is the angle between the direction of the force and the direction of motion
In our problem,
![F=5200 N](https://img.qammunity.org/2019/formulas/physics/high-school/qm9sgo4cpsf0mtzbqos21pvv6ijvzlcuyq.png)
,
![d=25 m](https://img.qammunity.org/2019/formulas/physics/high-school/e1r09bxwf4teow2aocwrahk606pc92nyas.png)
and
![\theta=0](https://img.qammunity.org/2019/formulas/physics/high-school/14jwpivphnfdvk8vrtofpnqklq15pddk29.png)
(because the force is parallel to the direction of motion), so the work done by the crane is
![W=(5200 N)(25 m)(\cos 0^(\circ))=1.3 \cdot 10^5 J](https://img.qammunity.org/2019/formulas/physics/high-school/mv2puelhsfe7w4ntgupjralcu13oqfk6ba.png)