Question

A machine carries a 3.0 kg package from an initial position of i = (0.50 m) + (0.75 m) + (0.20 m) at t = 0 to a final position of f = (7.80 m) + (12.5 m) + (7.40 m) at t = 11 s. The constant force applied by the machine on the package is = (2.00 N) + (4.00 N) + (6.00 N).

Answer 1

Answer:

mass of package, m = 3 kg

Step-by-step explanation:

initial position,
`\overrightarrow{r_(1)}=0.5\widehat{i}+0.75\widehat{j}+0.2\widehat{k}`

Final position,

`\overrightarrow{r_(2)}=7.8\widehat{i}+12.5\widehat{j}+7.4\widehat{k}`

time, t = 11 s

Force,

`\overrightarrow{F}=2\widehat{i}+4\widehat{j}+6\widehat{k}`

Work done,

`W = \overrightarrow{F}.\overrightarrow{d}`

where, d = r2 - r1

`\overrightarrow{d}=\left ( 7.8-0.5 \right )\widehat{i}+\left ( 12.5-0.75 \right )\widehat{j}+\left ( 7.4-0.2 \right )\widehat{k}`

`\overrightarrow{d}=7.3\widehat{i}+11.75\widehat{j}+7.2\widehat{k}`

W =
`\left (  2\widehat{i}+4\widehat{j}+6\widehat{k}\right ).\left (\overrightarrow{d}=7.3\widehat{i}+11.75\widehat{j}+7.2\widehat{k}  \right )`

W = 104.8 J

Power = Work / time

P = 104.8 / 11 = 9.53 Watt

Step-by-step explanation: