Respuesta :
The work is the difference between the final and initial kinetic energy:
[tex]W=\Delta K = \dfrac{1}{2}mv^2_f-\dfrac{1}{2}mv^2_i=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]
Substitute your values to get
[tex]W=\dfrac{1}{2}\cdot 2(25-0)=25[/tex]
Answer : The work done is, 25 J
Step-by-step explanation :
As we know that work is the difference between the final and initial kinetic energy.
That means,
[tex]w=K.E=\frac{1}{2}m\times v^2[/tex]
[tex]w=K.E=\frac{1}{2}m\times (v_f-v_i)^2[/tex]
where,
w = work done
m = mass = 2 kg
K.E = kinetic energy
[tex]v_i[/tex] = initial speed = 0 m/s
[tex]v_f[/tex] = final speed = 5 m/s
Now put all the given values in the above formula, we get:
[tex]w=\frac{1}{2}\times (2kg)\times (5-0)^2m^2/s^2[/tex]
[tex]w=25kg.m^2/s^2=25J[/tex]
Therefore, the work done is, 25 J
