A 398-kg boat is sailing 14.0° north of east at a speed of 1.50 m/s. 23.0 s later, it is sailing 33.0° north of east at a speed of 4.40 m/s. During this time, three forces act on the boat: a 32.8-N force directed 14.0° north of east (due to an auxiliary engine), a 20.5-N force directed 14.0° south of west (resistance due to the water), and (due to the wind). Find the (a) the magnitude and (b) direction of the force. Express the direction as an angle with respect to due east.

Question

Physics

Simeon Weiss

25 October, 13:35

A 398-kg boat is sailing 14.0° north of east at a speed of 1.50 m/s. 23.0 s later, it is sailing 33.0° north of east at a speed of 4.40 m/s. During this time, three forces act on the boat: a 32.8-N force directed 14.0° north of east (due to an auxiliary engine), a 20.5-N force directed 14.0° south of west (resistance due to the water), and (due to the wind). Find the (a) the magnitude and (b) direction of the force. Express the direction as an angle with respect to due east.

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Jasiah Pham · Answer 1

We shall represent speed in vector form

First speed

v₁ = 1.5 cos 14 i + 1.5 sin 14 j

= 1.455 i + 0.363 j

v₂ = 4.4 cos 33 i + 4.4 sin 33 j

= 3.69 i + 2.39 j

v₂ - v₁

3.69 i + 2.39 j - 1.455 i - 0.363 j

= 2.235 i + 2.027 j

acceleration

= v₂ - v₁ / time

= (2.235 i + 2.027 j) / 23

=.097 i +.088 j

force = mass x acceleration

= 398 x (.097 i +.088 j)

= 38.6 i + 35.02 j

Magnitude of force F

F² = 38.6² + 35.02²

F = 52.11 N

Tan θ = 35.02 / 38.6

θ = 42° north of east.