Question

After the box comes to rest at position x₁, a person starts pushing the box, giving it a speed v₁. When the box reaches position x₂ (where x₂>x₁), how much work $W_p$ has the person done on the box? Assume that the box reaches x₂ after the person has accelerated it from rest to speed v₁. Express the work in terms of m, v₀, x₁, x₂, and v₁.

Answer 1

Answer:

$W_p = \frac{1}{2}mv_1^2$

Explanation:

As we know that box is initially at rest

So we will have

$v_i = 0$

now as it will be displaced from initial position to final position then final speed of the box is reached to

$v_f = v_1$

now we know by work energy theorem that work done by all the forces on the box will be equal to the change in kinetic energy

So here we will have

$W_p = \frac{1}{2}m(v_f^2 - v_i^2)$

$W_p = \frac{1}{2}m(v_1^2 - 0)$

$W_p = \frac{1}{2}mv_1^2$