Question

A particle of mass 500 gm is moving in a straight line with velocity v = bx 5/2. 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).

• A. 2 J
• B. 4 J
• C. 8 J
• D. 16 J

Answer 1

Answer: (D)

16 J

Answer 2

The correct option is(D): 16 J.

W total = Δ K

$=\frac{1}{2}(\frac{1}{2})[{b(4)^{\frac{5}{2}}}^2-0]$

$=\frac{b^2}{4}×4^5$

⇒ W total = 16 J