Question

A particle is moving along x-axis with x as its position along x-axis. Its acceleration is given by a = 2x – 4. The work done by all forces on the particle between x =2 m to x = 4m is (mass of the particle is 1 kg)

Answer 1

Answer:

here

a=2x-4

or,v(dv/dx)=2x-4

or vdv=(2x-4)dx

integrating on voth sides at x=2 v=v1(say) and at x=4 v=v2 say

$\int\limits^{v_2}_{v_1} {vdv} =\int\limits^4_2 {(2x-4)} \, dx$

$\frac{1}{2} ({v_2}^{2} -{v_1^{2} )=4$..................(1)

the total work done is given by change in kinetic energy

ΔKE=$\frac{1}{2}m ({v_2}^{2} -{v_1^{2} )$

       =4*1

       =4J