An iceboat is at rest on a frictionless frozen lake when a sudden wind exerts a constant force of 200. N, towards the east, on the boat. Due to the angle of the sail, the wind causes the boat to slide in a straight line for a distance of 8.0 m in a direction 20 degrees north of east. What is the kinetic energy of the iceboat at the end of that 8.0 m?

If you looked at this and thought, "bro what the heck 1/2mv^2 doesnt work",

you're right. The kinetic energy formula doesn't work directly.

However, remember that the change in kinetic energy is equal to work. If our initial spot is at rest, that means we only care about the final.

Here's the equation for change in KE: (1/2 m*vf^2) - (1/2 m*vi^2)

Ignore initial because the boat is at rest -- vi = 0

change in KE is now : (1/2 m*vf^2)

Now remember: change in KE = Work

Work = F*d*cos(theta)

Plug in 20 degrees for theta, plug in 200N for force, and 8m for D.

Work = cos(20) * 200N * 8m

Work = 1503.508 = KE

Work ~ 1500 = KE

KE ~ 1500

are there options? cause you ain’t. real specific

gravity

explanation: