Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force [tex]F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k[/tex]
Mass of object = 2.00 kg
Initial position [tex]d=(2.75)i-(2.00)j+(5.00)k[/tex]
Final position [tex]d=-(5.00)i+(4.50)j+(7.00)k[/tex]
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done
[tex]W=F\cdot d[/tex]
[tex]W=F\cdot(d_{f}-d_{i})[/tex]
Put the value into the formula
[tex]W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)[/tex]
[tex]W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)[/tex]
[tex]W=2.75\times-7.75+7.50\times6.50+6.75\times2.00[/tex]
[tex]W=40.93\ J[/tex]
Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.
