Question

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitude of 6 000 N directed toward the northwest (45 N of W). What is the magnitude of the resultant acceleration? with steps, please.

Answer 1

The magnitude of the resultant acceleration is 2.2
`m/s^2`

Step-by-step explanation:

Mass (m) of the sailboat = 2000 kg

Force acting on the sailboat due to ocean tide is
`F_1` = 3000N

Eastwards means takes place along the positive x direction

Then
`F_(1x)` = 3000N and
`F_(1y)`= 0

Wind Force acting on the Sailboat is
`F_2` = 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis

Then

`F_(2x)` = -(6000N) cos 45 degree = -4242.6 N

`F_(2y)` = (6000N) cos 45 degree = 4242.6 N

Hence , the net force acting on the sailboat in x direction is

`F_x = F_(1x)+ F_(2x)`

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is

`F_y = F_(1y)+ F_(2y)`

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

`\sqrt{(1242.6)^2 + (4242.6)^2`

`√((1544054.76) + (17999654.8))`

`√((19543709.5)^2)`

4420.8 N

F = ma

`a = (F)/(m)`

`a =(4420.8)/( 2000)`

=2.2
`m/s^2`