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 = (1)/(2)mv^2_f-(1)/(2)mv^2_i=(1)/(2)m(v^2_f-v^2_i)`

Substitute your values to get

`W=(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=(1)/(2)m* v^2`

`w=K.E=(1)/(2)m* (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=(1)/(2)* (2kg)* (5-0)^2m^2/s^2`

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

Therefore, the work done is, 25 J