. An automobile is traveling at 90 km=hr around a curve of radius 375 m. If the

Question

. An automobile is traveling at 90 km=hr around a curve of radius 375 m. If the road is áat, what minimum coe¢ cient of friction is needed in order not to skid? a. 0:61 b. 0:45 c. 0:28 d. 0:34 e. 0:17

6. A spacecraft orbits an unknown planet at a distance of 5:2 107 m from its center. The period of its orbit is 52 hours. What is the mass of the planet? a. 2:4 1024 kg b. 3:5 1022 kg c. 9:2 1025 kg d. 6:1 1026 kg e. 1:9 1020 kg

7. A 64 kg box is pushed at constant speed up a frictionless plane for a distance of 18 m by a force parallel to the plane. If the plane makes an angle of 240 above the horizontal, how much work was done? a. 1500 J b. 1200 J c. 4600 J d. 2900 J e. 8800 J

8. A 10:0 kg mass starts at the top of an incline that is 8:0 m long and makes an angle of 400 above the horizontal. As it slides down, there is a constant frictional force of 23:8 N. What is its speed at the bottom? a. 6:1 m=s 1 b. 10:4 m=s c. 5:5 m=s d. 7:9 m=s e. 9:0 m=s

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Explanation / Answer

1) coefficient of friction=v2/rg=252/(375*9.8)=0.17 (Option (e))

2) speed V = r w = r x 2 pi / T = 2x3.14 x 5.2 x10^7/(52x3600) = 1.7453 km/sec

speed v = square root of ( G M / r), where M is mass of the planet

so that M = v^2 x r / G = 1745.3 x1745.3 x 5.2 x10^7 / 6.67 x10^-11 = 2.37 x10^24 Kg (Option (a))

3) W= mgsin24 *18 =64*9.8*18*0.4067= 4591 J (Option C)

4) motive force F = mgsin = 10kg * 9.8m/s² * sin40º = 63 N (downslope)
friction force Ff =23.8 N (upslope, against motion)
net force Fnet = (63 - 23.8)N = 39.2 N downslope
acceleration a = Fnet / m - 39.2N / 10kg = 3.92 m/s²
speed v = (2as) = (2 * 3.92m/s² * 8m) = 7.9 m/s (Option d)

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