Name: Just as a car tops a 36 meter high hill with a speed of 42 km/h it runs ou

Question

Name: Just as a car tops a 36 meter high hill with a speed of 42 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill? A 2 kg metal plate slides down a 16-meter high slope. At the bottom its speed is 4.3 m/s. To the nearest Joule, what was the magnitude of the work done by friction? A mass of 3.3 kg is dropped from a height of 3.86 meters above a vertical spring anchored at its lower end to the floor. If the spring constant is 34 N/cm, how far, to the nearest tenth of a cm, is the spring compressed? If the top of the spring in the preceding problem is 1.83 meters above the ground when the mass is released, what is the ball's kinetic energy, to the nearest Joule, just before the mass strikes the spring? A boy on a bicycle drags a wagon full of newspapers at 0.800 m/s for 30.0 min using a force of 40.0 N. How much work has the boy done?

Explanation / Answer

1) On the next high all the energy will be potential (m*g*h)
So m*g*h2 = m*g*h1 + 1/2*m*v^2 ..........(U + K) potential + kinetic

So mass drops out leaving
h2 = h1 + v^2/2g = 36m + (42km/hr*(1000m/km)*(1hr/3600s))^2/(2*9.8) = 42.94 m

2)  U + W = K ...so W = K - U = 1/2*m*v^2 - m*g*h = 1/2*2kg*4.3^2 - 2*9.8*16 = -295.11 J

3) The height energy lost goes into spring compression energy
Height energy lost = 3.3*9.8*(3.86 + x)
Spring compression energy = 0.5*3400*x^2
1700 x^2 = 32.34 x + 124.83

x = 28.07 cm

5) v = 0.8m/s = 48 m/min

d = v*t = 48*30 = 1440 m

F = 40 N

Work = W = F*d = 40*1440 = 5.76*10^4 J

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