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horsena [70]
3 years ago
5

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.
Physics
1 answer:
Natasha2012 [34]3 years ago
4 0

Answer:

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

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\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

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

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