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3 PHY QUESTIONS about force
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1)A 3.48kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by s(t)=(1.80m/s)t +(2.11m/s^3 )t^3 What is the magnitude of the force F when 6.20s ?... 顯示更多 1)A 3.48kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by s(t)=(1.80m/s)t +(2.11m/s^3 )t^3 What is the magnitude of the force F when 6.20s ? Express your answer with the appropriate units. 2)A binary star system consists of two stars of masses m 1 and m 2 . The stars, which gravitationally attract each other, revolve around the center of mass of the system. The star with mass m 1 has a centripetal acceleration of magnitude a 1 . Note that you do not need to understand universal gravitation to solve this problem. Find a 2 , the magnitude of the centripetal acceleration of the star with mass m 2 . Express the acceleration in terms of quantities given in the problem introduction. 3)A gymnast of mass 61.0kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81m/s 2 for the acceleration of gravity. a)Calculate the tension T in the rope if the gymnast hangs motionless on the rope. b)Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate. c)Calculate the tension T in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 0.800m/s^2 . d)Calculate the tension T in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 0.800m/s^2 Express all answer in newtons. . THX a lot .
最佳解答:
1. velocity v = ds/dt = 1.8 + 6.33t^2 acceleration a = dv/dt = 12.66t Hence, acceleration at t= 6.2 s equals to 12.66 x 6.2 m/s^2 = 78.49 m/s^2 Therefore, F - 3.48g = 3.48 x 78.49 F = 307 N (take g = 9.81 m/s^2) 2. Centripetal force on m1 = (m1).(a1) centripetal force on m2 = (m2)(a2) Since centripetal force is provided by the same gravitational attractive force, thus (m1)(a1) = (m2)(a2) i.e. a2 = (m1/m2).(a1) 3. (a) Tension = weight of gymnast = 61g N = 598 N (b) same as in (a). (c) Force that the gymnast needs to exert on the rope = (598 + 61 x 0.8) N = 647 N Tension in rope = 647 N (d) Force that the gymnast needs to exert on the rope = (598 - 61 x 0.8) N = 549 N Tension in rope = 549 N
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