Question

A particle is moving along the x-axis. It starts from rest at x = 0 and moves to x = L under the action of a variable force with an x-component given by Fx(x): F(x) x2 if 0 SX SL 1 L2 -=Lx if L < x < 3L 2 2 Everything is in SI units: x in meters and F in Newtons. 1: What is the particle's kinetic energy at x = L? Express your answer in terms of L and Fo. K (x = 1/2) = 2: What is the particle's kinetic energy at x = 3L? Express your answer in terms of L and Fo. K (x = L) = 3: If the mass of the particle is 2 kg and L = 3 meters, what is the speed of the particle at x = 3L? V=

Answer 1

Kinetic energy is defined as the work done in moving a particle under a force so

$K =\int_{0}^{x}Fdx$

So,

ques 1 K.E at x = l

$K =\int_{0}^{L}Fdx$

$K =\int_{0}^{L}F_0 x^2dx$

$K =\frac{F_0}{3}L^3$

Ques 2 K.E = 3L

$K =\int_{0}^{3L}Fdx$

$K =F_0(\int_{0}^{L}x^2dx + \int_{L}^{3L}\left ( \frac{3}{2}L^2 - \frac{1}{2}Lx \right )dx)$

$K =F_0(\frac{L^3}{3} + \left ( 3L^3 - 2L^3 \right ))$

$K =F_0\frac{4L^3}{3}$

ques 3

m= 2kg

L = 3 m

So

$K =F_0\frac{4L^3}{3} = \frac{1}{2}mv^2$

here F₀ = 1

$K =\frac{4(3)^3}{3} = \frac{1}{2}(2)v^2$