ooo T-Mobile LTE * 36%. \" 11:25 PM webassign.net A 2.3 kg block is dropped from
ID: 3280765 • Letter: O
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ooo T-Mobile LTE * 36%. " 11:25 PM webassign.net A 2.3 kg block is dropped from rest from a height of 4.6 m above the top of the spring. When the block is momentarily at rest, the spring is compressed by 25.0 cm. What is the speed of the block when the compression of the spring is 15.0 cm? Let the system include the block, the spring, and Earth. Let the gravitational potential energy be zero where the spring is compressed by the given distance. Then the mechanical energy at this compression distance will be partially kinetic and partially stored in the spring. You can use conservation of mechanical energy to relate the initial potential energy of the system to the energy stored in the spring and the kinetic energy of block when it has compressed the spring the given distance. In a volcanic eruption, 3.0 km3 of mountain with a density of 1600 kg/m3 was lifted an average height of 490 m. (a) How much energy in joules was released in this eruption? (b) The energy released by thermonuclear bombs is measured in megatons of TNT, where 1 megaton of TNT = 4.2. 1015 . Convert your answer for (a) to megatons of TNT Mton TNT A 2000 kg car moving at an initial speed of 30 m/s along a horizontal road skids to a stop in 70 m. (Note: When stopping without skidding and using conventional brakes, 100 percent of the kinetic energy is dissipated by friction within the brakes. With regenerative braking, such as that used in hybrid vehicles, only 70 percent of the kinetic energy is dissipated.) (a) Find the energy dissipated by friction. ko) (b) Find the coefficient of kinetic friction between the tires and the road.Explanation / Answer
Solving first question
Conserving energy
Initial energy=final energy
Mgh=0.5*m*v*v+0.5*k*x*x........(1)
We need value of k now
Conserving energy between final compression and initial point
2.3*9.8*4.6=0.5*k*0.25*0.25
We got k=3317.9 N/m
Putting in (1) we got
v=7.6 m/s
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