In Problem 21 it is observed that when the velocity of the sailboat reaches 5 m/sec, the boat begins to rise out of the water and “plane.” When this happens, the proportionality constant for the water resistance drops to b0 = 60 N-sec/m. Now find the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind as it is planning?
There are two forces are acting in the direction of the boat and in the opposite direction of the boat respectively.
Put v = 5, t = 0 then k = 6
Hence, The equation of motion is .
The limiting velocity of the sailboat is
Hence, the Limiting velocity is 10m/sec.
When an object slides on a surface, it encounters a resistance force called friction. This force has a magnitude of , where the coefficient of kinetic friction and N is the magnitude of the normal force that the surface applies to the object. Suppose an object of mass 30 kg is released from the top of an inclined plane that is inclined to the horizontal (see Figure 3.11). Assume the gravitational force is constant, air resistance is negligible, and the coefficient of kinetic friction . Determine the equation of motion for the object as it slides down the plane. If the top surface of the plane is 5 m long, what is the velocity of the object when it reaches the bottom?
When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v2 with the force acting in opposition to the motion of the object. A shell of mass 3 kg is shot upward from the ground with an initial velocity of 500 m/sec. If the magnitude of the force due to air resistance is 0.1v2, when will the shell reach its maximum height above the ground? What is the maximum height?
During the summer the temperature inside a van reaches , while that outside is a constant . When the driver gets into the van, she turns on the air conditioner with the thermostat set at . If the time constant for the van is and that for the van with its air conditioning system is , when will the temperature inside the van reach ?
Rocket Flight. A model rocket having initial mass mo kg is launched vertically from the ground. The rocket expels gas at a constant rate of a kg/sec and at a constant velocity of b m/sec relative to the rocket. Assume that the magnitude of the gravitational force is proportional to the mass with proportionality constant g. Because the mass is not constant, Newton’s second law leads to the equation (mo - αt) dv/dt - αβ = -g(m0 – αt), where v = dx/dt is the velocity of the rocket, x is its height above the ground, and m0 - αt is the mass of the rocket at t sec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 0≤t<m0/α.
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