Question(II) A box of mass 4.0 kg is accelerated from rest by a force across a floor at a rate of $2.0 \mathrm{~m} / \mathrm{s}^{2}$ for 8.0 s . Find the net work done on the box.

Studdy Solution

STEP 1

1. The box has a mass of $4.0 \, \text{kg}$ .
2. The box starts from rest.
3. The box is accelerated at a rate of $2.0 \, \text{m/s}^2$ .
4. The acceleration occurs over a time period of $8.0 \, \text{s}$ .
5. We need to find the net work done on the box.

STEP 2

1. Calculate the final velocity of the box.
2. Determine the displacement of the box.
3. Calculate the net work done using the work-energy principle.

STEP 3

Calculate the final velocity of the box using the formula for acceleration:
$v = u + at$
where $u = 0 \, \text{m/s}$ (initial velocity), $a = 2.0 \, \text{m/s}^2$ (acceleration), and $t = 8.0 \, \text{s}$ (time).
$v = 0 + (2.0 \, \text{m/s}^2)(8.0 \, \text{s})$ $v = 16.0 \, \text{m/s}$

STEP 4

Determine the displacement of the box using the formula:
$s = ut + \frac{1}{2}at^2$
Substitute $u = 0 \, \text{m/s}$ , $a = 2.0 \, \text{m/s}^2$ , and $t = 8.0 \, \text{s}$ :
$s = 0 + \frac{1}{2}(2.0 \, \text{m/s}^2)(8.0 \, \text{s})^2$ $s = \frac{1}{2}(2.0)(64.0)$ $s = 64.0 \, \text{m}$

STEP 5

Calculate the net work done using the work-energy principle:
$W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$
where $m = 4.0 \, \text{kg}$ , $v = 16.0 \, \text{m/s}$ , and $u = 0 \, \text{m/s}$ .
$W_{\text{net}} = \frac{1}{2}(4.0 \, \text{kg})(16.0 \, \text{m/s})^2 - \frac{1}{2}(4.0 \, \text{kg})(0)^2$ $W_{\text{net}} = \frac{1}{2}(4.0)(256.0)$ $W_{\text{net}} = 512.0 \, \text{J}$
The net work done on the box is:
$\boxed{512.0 \, \text{J}}$

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