Question

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m)i − (2.00 m)j + (5.00 m)k to a final position of d = −(5.00 m)i + (4.50 m)j + (7.00 m)k in 4.00 s. Find the work done on the object by the force in the 5.60 s interval.

Answer 1

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Step-by-step explanation:

Given that,

Force
`F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k`

Mass of object = 2.00 kg

Initial position
`d=(2.75)i-(2.00)j+(5.00)k`

Final position
`d=-(5.00)i+(4.50)j+(7.00)k`

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

`W=F\cdot d`

`W=F\cdot(d_(f)-d_(i))`

Put the value into the formula

`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)`

`W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)`

`W=2.75*-7.75+7.50*6.50+6.75*2.00`

`W=40.93\ J`

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.