Question

Force acting on a body of mass 1 kg is related to its position x as F = (x 3 – 3x) N. It is at rest at x = 1.Its velocity at x = 3 can be :

Answer 1

Given : Force acting on a body of mass 1kg is related to its position x as F = (x³ - 3x) N. body is at rest at x = 1.

To find : The velocity of body at x = 3 can be ..

solution : F = (x³ - 3x) N

from Newton's 2nd law, F = ma

⇒ma = (x³ - 3x)

⇒1kg × a = (x³ - 3x) [ mass of body, m = 1kg ]

⇒a = (x³ - 3x)

acceleration is the rate of change of velocity with respect to time.

i.e., a = dv/dt = v dv/dx

⇒vdv/dx = (x³ - 3x)

⇒$\int\limits^v_u vdv=\int\limits^3_1{x^3-3x}\,dx$

⇒(v² - u²)/2 = $\left[\frac{x^4}{4}-\frac{3}{2}x^2\right]^3_1$

as body is at rest at x = 1, so u = 0

⇒v²/2 - 0 = [3⁴/4 - 3/2 (3)² ] - [1⁴/4 - 3/2 (1)²]

⇒v²/2 = [20.25 - 13.5 ] - [0.25 - 1.5 ]

⇒v²/2 = 6.75 + 1.25 = 8

⇒v² = 16

⇒v = 4

Therefore the velocity of the body at x = 3 can be 4 m/s.