Question

A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N. The sailboat travels the distance in 1.0 h. How much work was done by the wind? What was the winds power? Must show all work A sailor pushes a 195.0 kg crate up a ramp that is 2.00 m high and 6.00 m long onto the deck of a ship. He exerts a 750.0 N force parallel to the ramp. What is the mechanical advantage of the ramp? what is the efficiency of the ramp? Must show all work A man lifts various loads with the same lever. The distance of the applied force from the fulcrum is 2.50 m and the distance from the fulcrum to the load is 0.500 m. A graph of resistance force vs. effort force s shown. What is the mechanical advantage of the lever? What is the ideal mechanical advantage of the lever? what is the efficiency of the lever? Show all work A girl places a stick at an angle of 60.0 degrees against a flat rock on a frozen pond. She pushes at an angle and moves the rock horizontally for 3.00 m across the pond at a velocity of 5.00 m/s and a power of 250.0 W. What force did she apply to the stick? How much work did she do? Show all work

Answer 1

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 $45^{\circ}$.

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

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
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

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:
$MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55$

Finally, the efficiency $\epsilon$ 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:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

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:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is 
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$