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Math  /  Algebra

Question1) A 10.31−kg10.31-\mathrm{kg} box is at rest when Joseph begins pushing it with a force of +18.66 N , causing it to move 3.64 m . Neglect friction. a) How much work did Joseph do on the box? \begin{tabular}{|l|l|} \hline & J \\ \hline \end{tabular} b) How fast is the box moving after 3.64 m ? \begin{tabular}{|ll|} \hline 3.63 m/s3.63 \mathrm{~m} / \mathrm{s} \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. The box has a mass of 10.31 kg.
2. Joseph applies a force of 18.66 N.
3. The box moves a distance of 3.64 m.
4. Friction is neglected.
5. We need to calculate the work done by Joseph on the box.
6. We need to calculate the speed of the box after moving 3.64 m.

STEP 2

1. Calculate the work done by Joseph on the box.
2. Calculate the final speed of the box after moving 3.64 m.

STEP 3

Calculate the work done by Joseph on the box using the formula for work, which is the product of force and displacement in the direction of the force:
W=F×d W = F \times d
where F=18.66 N F = 18.66 \, \text{N} and d=3.64 m d = 3.64 \, \text{m} .
W=18.66 N×3.64 m W = 18.66 \, \text{N} \times 3.64 \, \text{m}

STEP 4

Calculate the numerical value:
W=67.9704 J W = 67.9704 \, \text{J}
Joseph did 67.97 J \boxed{67.97} \, \text{J} of work on the box.

STEP 5

Calculate the final speed of the box using the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy:
W=ΔKE=12mv2−12mu2 W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2
Since the box starts from rest, u=0 u = 0 , so:
W=12mv2 W = \frac{1}{2} m v^2
Rearrange to solve for v v :
v2=2Wm v^2 = \frac{2W}{m}
v=2Wm v = \sqrt{\frac{2W}{m}}
where W=67.9704 J W = 67.9704 \, \text{J} and m=10.31 kg m = 10.31 \, \text{kg} .

STEP 6

Substitute the values and calculate:
v=2×67.9704 J10.31 kg v = \sqrt{\frac{2 \times 67.9704 \, \text{J}}{10.31 \, \text{kg}}}
v=135.940810.31 v = \sqrt{\frac{135.9408}{10.31}}
v=13.186 v = \sqrt{13.186}
v≈3.63 m/s v \approx 3.63 \, \text{m/s}
The box is moving at 3.63 m/s \boxed{3.63} \, \text{m/s} after 3.64 m.

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