**The Work Done in Pulling a Supertanker**

Part A = 1.65 * 10^9 J

**Solution Below:**

^{6}N, one at an angle 13.0° west of north, and the other at an angle 13.0° east of north, as they pull the tanker a distance 0.650 km toward the north.

Part A

Express your answer in joules, to three significant figures.

Work is force times distance. The tugs are pulling at an angle to north, but each is pulling in the opposite direction (with respect to the x-direction) – so the forces cancel out in the x-direction, and the only force left is acting in the y-direction. To begin, find the y-component of the force from one of the tugs and then double it:

Fy = F*cos(θ)

Fy = 1300000 * cos(13)

Fy = 1266681 N

Now double the force, since there are two tugs. The overall force is 2533362 N. Now multiply by the distance. 0.650 km is 650m:

W = FD

W = 2533362 * 650

W = 1646685409 J

Rounded to 3 significant figures, this is 1650000000 J, or 1.65 * 10^9

1.65 * 10^9

Part B

Express your answer in joules, to three significant figures.

Work is force times distance. The tugs are pulling at an angle to north, but each is pulling in the opposite direction (with respect to the x-direction) – so the forces cancel out in the x-direction, and the only force left is acting in the y-direction. To begin, find the y-component of the force from one of the tugs and then double it:

Fy = F*cos(θ)

Fy = 1300000 * cos(13)

Fy = 1266681 N

Now double the force, since there are two tugs. The overall force is 2533362 N. Now multiply by the distance. 0.650 km is 650m:

W = FD

W = 2533362 * 650

W = 1646685409 J

Rounded to 3 significant figures, this is 1650000000 J, or 1.65 * 10^9 J

1.65 * 10^9 J

to get y component of a force you use sin not cosine

This is true, however the problem gives the angle with respect to the y-axis (west of north, instead of north of west)