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 wp 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

The work done by the person on the box can be calculated using the work-energy theorem. It is equal to the change in kinetic energy of the box, which can be expressed in terms of the mass and speed of the box.

Step-by-step explanation:

When the box is pushed from rest to speed v1, the person does work on the box. This work, wp, can be calculated using the work-energy theorem. The work done is equal to the change in kinetic energy of the box, which is given by the equation:

wp = (1/2)mv₁² - (1/2)mv₀²

where m is the mass of the box, v₁ is the final speed of the box, and v₀ is the initial speed of the box (which is zero since it starts from rest).

So the work done by the person on the box can be expressed in terms of m, v₀, x₁, x₂, and v₁ as:

wp = (1/2)m(v₁² - v₀²)

Answer 2

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

`W_p = KE_f - KE_i`

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

`KE_f = (1)/(2)mv_1^2`

now work done is given as

`W_p = (1)/(2)mv_1^2 - 0`

so we can say

`W_p = (1)/(2)mv_1^2`

so above is the work done on the box to slide it from x1 to x2