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Question: An object is acted upon by the forces \(\vec { F } _ { 1 }\)= \(4 \hat { i }\) N and \(\vec { F } _...

An object is acted upon by the forces F1\vec { F } _ { 1 }= 4i^4 \hat { i } N and F2\vec { F } _ { 2 } =

( i^j^\hat { \boldsymbol { i } } - \hat { \boldsymbol { j } } )N. If the displacement of the object is D =

( i^+6j^6k^\hat { i } + 6 \hat { j } - 6 \hat { k } )m, the kinetic energy of the object

A

Remains constant

B

Increases by 1J

C

Decreases by 1 J

D

Decreases by 2J

Answer

Decreases by 1 J

Explanation

Solution

The work done on the object W = F1D+F2D\vec { F } _ { 1 } \cdot \vec { D } + \vec { F } _ { 2 } \cdot \vec { D } where is the displacement vector.

∴ W = 4(1) + (1) (1) + (−1) (6) = −1J

From work energy theorem W = KEf – KEi = −1J

⇒ Kinetic energy decrease by 1J. Hence, the correct choice is (3)