Question

ANSWER QUICKLY PLEASE

Answer 1

-mg(h₂ - h₁)

Step-by-step explanation:

The total work done on a particle as it moves from positions h₁ to h₂ in the 2 direction can be determined by evaluating the integral of the force over the displacement. In this case, the force is given by vec F = -mg hat z.

Using the integral definition of work, we have:

W = ∫(r₁ to r₂) (vec F ⋅ dvec r)

Since the displacement is in the 2 direction, the position vector can be written as dvec r = dz hat z.

Substituting the force and displacement, we have:

W = ∫(h₁ to h₂) (-mg hat z ⋅ dz hat z)

Since hat z ⋅ hat z = 1, the dot product simplifies to -mg.

W = -mg ∫(h₁ to h₂) dz

Integrating with respect to z, we get:

W = -mg [z] evaluated from h₁ to h₂

W = -mg (h₂ - h₁)

Therefore, the total work done on the particle as it moves from positions h₁ to h₂ in the 2 direction is -mg(h₂ - h₁).

Answer 2

The last option,
`W=-mg(h_2-h_1)`

Step-by-step explanation:

Given a force and displacement vector, we are asked to find the total work done on a particle as it moves from positions h₁ to h₂ in the z direction.

The integral definition of work is written as:

`\rightarrow W=\displaystyle \int\limits^(r_2)_(r_1) {\vec F \cdot} \, d\vec r`

Given:

`\vec F=\big < 0,0,-mg\big > \\\\d\vec r=\big < dx,dy,dz \big >`

Substituting our given values into the work definition:

`\Longrightarrow W=\displaystyle \int\limits^(h_2)_(h_1) {\big < 0,0,-mg\big > \cdot} \, \big < dx,dy,dz \big >`

Taking the dot product of what's in the integrand:

`\Longrightarrow W=\displaystyle \int\limits^(h_2)_(h_1) {\big(-mg\big) } \, dz`

Now evaluating the integral with respect to "z":

`\Longrightarrow W=-mgz\Big|\limits^(h_2)_(h_1)\\\\\\\\\Longrightarrow W= (-mg(h_2))-(-mg(h_1))\\\\\\\\\Longrightarrow W= mgh_1-mgh_2\\\\\\\\\Longrightarrow W= mg(h_1-h_2)\\\\\\ \\ \Longrightarrow W= -mg(-h_1+h_2) \\\\\\\\\therefore \boxed{\boxed{W=-mg(h_2-h_1)}}`

Therefore, the last option is correct.