Question
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= 4i^ N and F2 =
( i^−j^ )N. If the displacement of the object is D =
( i^+6j^−6k^ )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 = F1⋅D+F2⋅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)