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3. An apparatus for launching a small boat consists of a 150.0kg cart that rides down a set of tracks on four solid steel wheels,each with radius 20.0 $\mathrm{cm}$ and mass 45.0 $\mathrm{kg}$ . The tracks slope at an angle of $7.50^{\circ}$ to the horizontal, and the boat's mass is 750.0kg. If the boat is released from rest a distance of 16.0 $\mathrm{m}$ fromthe water (measured along the slope), how fast will it be mov-ing when it reaches the water? Assume the wheels roll withoutslipping, and that there is no energy loss due to friction.

a) $\sqrt{2 g H}$b) $\sqrt{\frac{6 g H}{5}}$

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

Simon Fraser University

McMaster University

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in this programme, we're going to use the conservation of energy to solve for the final velocity of boat. So the form of the conservation of energy that we're going to use is this. The changing kinetic energy must be matched with a decrease in the potential energy as the boat was on Lorraine, Now a bell to you is equal to in GI Delta. Y well, I'm letting capital L b the combined mass of the whole system, so let's calculate that now. So we have 150 kilograms from the card itself. We have four times the mass of each wheel, which is 45 kilograms. What's the matter, Bill which is 7 50 So this combines to us 1,080 kilograms, Doctor, why is the high change of the boat and it is equal to negative 16 which is the length of the ramp. But we need the vertical component on Lee. So we need two multiplied by sine of some 20.5 degrees. And this gives negative 2.9 So the boat drops in height by a little over two meters throughout the course of this trip. Now the kinetic energy is equal to 1/2 times the total mass times the square. This is the translational part, plus four times the rotational part of each of the wheels, which is 1/2 I Omega Square now because the wheels roll without slipping. We have that omega is equal to V over R and I, for each of the wheels is that of a cylinder since wheel Khun B assumed to be cylinders, and this is equal to 1/2 R squared, so I'm letting little him here be the mass of each wheel and army, the radius of each wheel. And so let's go ahead and pauses the kinetic energy and focus on potential so this again is equal to Capitol Hill Times. T times, Delta Why and now I can plug in or total mass they found and little high change. Physical 2,080 times 9 20 times negative 2.9 And so the total potential energy lost by the boat is equal to negative. 2.21 times 10 For Jules, by conservation of energy, this must be the energy that is now kinetic that the boat has Andalus calculate the speed. So I'm going back here, except I'm going to plug in what eyes? I'm implying what omega is into the expression. When I do that, I get K is equal to 1/2 only square. So the translation part does not change plus four because their four wheels dies 1/2 and then I is 1/2 film are square and then omega is V over R in that square. I closed my bracket here. So then I'm just gonna plug in what these values are. It's equal to 1/2 times 1,080 plus 45 close bracket B square, which is equal to 5.85 b square. So this is the connect energy that the boat has. This is the potential energy of lost by conservation of energy. This right here, they must be equivalent. After we get rid of this minus sign and so 5 25 the sward must be equal to 2.21 instance of fourth. And this means that B is equal to 6.15 meters per second. And that is the violence

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