4) A box is sliding down an incline tilted at an angle 14.0 the incline at a spe

Question

4) A box is sliding down an incline tilted at an angle 14.0 the incline at a speed of 1.70 m/s is0.380 How far does the box slide down the incline before coming to rest? horizontal. The box is sliding down The coefficient of kinetic friction between the box and the incline the A) 1.78 m B) 0.720 m C) 2.33 m D) 1.16 m E) The box does not stop. It accelerates down the plane g ball is moving with a constant speed of 5 m/s in a horizontal airdle whose What is the magnitude of the net force on the ball? A) 50 N B) 100 N C)ON D) 40 N E) 20N 6) List the four fundamental forces in nature. A) gravitational, electromagnetic, strong nuclear, weak nuclear B) gravitational, electromagnetic, contact, nuclear C) gravitational, contact, strong nuclear, weak nuclear D) gravitational, normal, kinetic friction, static friction E) gravitational, normal, tension, friction 7) Ordinary forces, such as pushes, pulls, normal forces and friction, are considered to be due to the A) gravitational force B) contact force. C) weak nuclear force. D) electromagnetic force. E) strong nuclear force. 8) Kepler's second law tells us that a planet sweepa out equal areas in equal times. If you compare the amount of area per time swept by Earth with the one of Jupiter, you would conclude A) They sweep the same area per time. B) Jupiter sweeps a larger area per time because it has much more mass than Earth C) Earth sweeps a larger area per time because it has much less mass than Jupiter D) Earth sweeps a larger area per time because it has a much smaller orbital path than Jupiter E) Jupiter sweeps a larger area per time because it has a much larger orbital path than Earth. 9) What is the force exerted by the Sun on the earth? The mass of the Sun is 199 x 100 kg, the mass of the earth is 597x 1024 kg, the Earth-Sun distance is 1.50 × 1011 m, and G-667-10-11 N m2/kg A) 3.52 x 1022N B)8.85 × 108 N C) 7.25x 1019 N D) 9.32 101l N E) 5.28 1033 N 2

Explanation / Answer

4) Fnet = m g sin 14 - (0.380 * m g cos 14)

= (m * 9.8 * sin 14 ) - (0.380 * m * 9.8 cos 14) = -1.24 m

a = F / m = -1.24 m / m = -1.24 m/s2

v2 - u2 = 2 a s

2 * -1.24 * s = -1.72

s = 1.72 / (2 * 1.24)

= 1.16 m [option D]

5) Force = m v2 / r

= 2 * 52 / 0.50

magnitude of net force = 100 N [option B]

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