Karen and Randy are playing with a toy car and track. They set up the track on t
ID: 1551711 • Letter: K
Question
Karen and Randy are playing with a toy car and track. They set up the track on the floor as shown in the figure below, where the apparatus on the far left is used to launch the car forward by pressing down on the top portion. After the car is launched, it follows the track and continues, leaving the track at an angle of theta = 41.2 degree with respect to the floor, at a height of h = 0.32 m above the floor. If the car has a mass of 124 g and it leaves the track with a speed of v = 2.56 m/s, what is the maximum height reached by the car, relative to the floor? Use conservation of energy to solve this problem. Ignore the effects of friction and air resistance. As of 2012, all of the shuttles in NASA's space shuttle program have been retired. When they once launched, a space shuttle (mass = 2.03 times 10^6 kg) would lift off as the thrust of its three main engines was suddenly augmented by the firing of the solid rocket boosters at t = 0 s. Suppose the magnitude of the shuttle's acceleration is given by a = 2.90t + 0.315t^2 - 0.105t^3, where a is in meters per second squared and t is in seconds, in the first few seconds after liftoff. What is the change in the kinetic energy of the space shuttle during the first 3.65 s after liftoff in this scenario?Explanation / Answer
velocity v = 2.56 m
angle with horizontal = 41.2 deg
vertical component of the vel vy = 2.56 Sin(41.2) = 1.69 m/s
PE at the launch = mgh
KE = mv2/2 = m*1.692/2 = 1.43 m
Total energy = mgh +1.43
when the car reached maximum height H then it will have only potentila energy mgH and KE =0 as its vertical velocity =0
conaserving the energy
mgH = mgh + m1.43
H = h +1.43/g = 0.32 +1.43/9.8 = 0.47 m
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