A particle of mass 500 gm is moving in a straight line with velocity v = bx5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m–3/2s–1).
- 2 J
- 4 J
- 8 J
- 16 J
The Correct Option is D
Solution and Explanation
The correct option is(D): 16 J.
Wtotal = ΔK
\(=\frac{1}{2}(\frac{1}{2})[{b(4)^{\frac{5}{2}}}^2-0]\)
\(=\frac{b^2}{4}×4^5\)
⇒ Wtotal = 16 J
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Concepts Used:
Relative Velocity
The velocity with which one object moves with respect to another object is the relative velocity of an object with respect to another. By relative velocity, we can further understand the time rate of change in the relative position of one object with respect to another.
It is generally used to describe the motion of moving boats through water, airplanes in the wind, etc. According to the person as an observer inside the object, we can compute the velocity very easily.
The velocity of the body A – the velocity of the body B = The relative velocity of A with respect to B
V_{AB} = V_{A} – V_{B}
Where,
The relative velocity of the body A with respect to the body B = V_{AB}
The velocity of the body A = V_{A}
The velocity of body B = V_{B}