Question

A 2-kg toy car accelerates from 0 to 5 m/s. How much work is done?

Answer 1

The work is the difference between the final and initial kinetic energy:

$W=\Delta K = \dfrac{1}{2}mv^2_f-\dfrac{1}{2}mv^2_i=\dfrac{1}{2}m(v^2_f-v^2_i)$

Substitute your values to get

$W=\dfrac{1}{2}\cdot 2(25-0)=25$

Answer 2

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,

$w=K.E=\frac{1}{2}m\times v^2$

$w=K.E=\frac{1}{2}m\times (v_f-v_i)^2$

where,

w = work done

m = mass = 2 kg

K.E = kinetic energy

$v_i$ = initial speed = 0 m/s

$v_f$ = final speed = 5 m/s

Now put all the given values in the above formula, we get:

$w=\frac{1}{2}\times (2kg)\times (5-0)^2m^2/s^2$

$w=25kg.m^2/s^2=25J$

Therefore, the work done is, 25 J