These are actually 4 different exercises:
ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is [tex]45^{\circ}[/tex].
Calling F the force of the wind, and [tex]d=15~km=15000~m[/tex] the distance covered by the boat, the work done by the wind is:
[tex]W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J[/tex]
The total time of the motion is [tex]t=1~h=3600~s[/tex] and therefore the power of the wind is
[tex]P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W [/tex]
ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
[tex]d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m [/tex]
The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
[tex]MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55 [/tex]
Finally, the efficiency [tex]\epsilon[/tex] of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
[tex]\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81 [/tex]
ex 3) the graph is missing
ex 4) We know that the power is the ratio between the work done W and the time t:
[tex]P= \frac{W}{t} [/tex]
But we can rewrite the work as
[tex]W=Fdcos\theta[/tex]
where F is the force applied, d the displacement of rock and [tex]\theta=60^{\circ][/tex] is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
[tex]P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta [/tex]
where [tex]v=d/t=5~m/s[/tex] is the velocity, Using the data of the exercise, we can then find the force, F:
[tex]F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N [/tex]
and now we can also calculate the work, which is
[tex]W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J[/tex]
