To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore
[tex]W = \Delta KE[/tex]
[tex]\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2[/tex]
Here,
m = mass
[tex]v_{f,i}[/tex] = Velocity (Final and initial)
First case) When the particle goes from 10m/s to 20m/s
[tex]\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2[/tex]
[tex]\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2[/tex]
[tex]W_1 = 150(m) J[/tex]
Second case) When the particle goes from 20m/s to 30m/s
[tex]\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2[/tex]
[tex]\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2[/tex]
[tex]W_1 = 250(m) J[/tex]
As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.