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I need these questions to have answers and to be worked out. A mass of 200 kg tr

ID: 3900267 • Letter: I

Question

I need these questions to have answers and to be worked out.

A mass of 200 kg travels down a frictionless roller coasterfrom a height of 50 m. When it reaches the level ground (height 0 m) it encounters an unstretched horizontal spring with spring constant 10,000 N/m. As the mass comes to rest how far does it compress the spring? A skateboarder rides his skateboard down the wall of a halfpipe from an initial height of 10 m. On the bottom of the otherwise frictionless halfpipe he encounters a region 10 m long that has a coefficient of friction 0.05. How high up the far side of the halfpipe does he rise before he comes to rest? A skier at the top of the 20 m tall 'bunny slope' skis down the frictionless slope to the flat ground below where she runs out of snow and reaches the parking lot, mu=2.0. How far across the parking lot does she travel before she comes to a stop? A man of mass 100 kg climbs a flight of stairs 20 m high in 120 s. How much power was required? In order for the mass traveling around a circular roller coaster loop to stay on the track at the top of the loop, what must the mass' minimum speed be at the top of the 10 m radius loop? A frictionless roller coaster car is released from rest on the track at a height of 50 m. How fast is the car going at the top of a 5 m radius circular loop?

Explanation / Answer

8) When the mass roll downs its potential energy will be converted to spring's kinetic energy
therefore mgh = 1/2kx^2
which gives x = (2mgh/k)^0.5
Putting values x = (2*200*9.8*50/10000)^0.5 = 4.42m

9)
Skateboarder's potential energy will be lost against friction
Initial energy = mgh = 10mg
Enerfy lost against friction = umgs =0.05*10mg = 0.5mg

Energy remaining = 10mg-0.5mg = 9.5mg
Now this will be converted to potential energy while going up therefore
9.5mg = mgH
therefore H = 9.5m

10) Skier's potential energy will be lost against friction therefore
mgh = umgs
therefore s = h/u = 20/2 = 10m

11) Power is the rate of work done therefore
Power = mgh/t = 100*20*9.8/120 = 163.33 W

12)sum of forces on an object = mass x accel of the object

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