speed. (a) What is the tension in the rope? (b) At a certain point the speed of
ID: 2031538 • Letter: S
Question
speed. (a) What is the tension in the rope? (b) At a certain point the speed of the bucket begins to change. The bucket now has an upward constant acceleration of 5. [18] A 6-kg bucket of water is being pulled straight up by a string at a constam magnitude 3 m/s. What is the tension in the rope now? (c) Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude 3 m/'s. Now what is the tension in the rope? 6. [12] A uniform cable of weight w hangs vertically downward, supported by an upward force of magnitude w at its top end. (a) What is the tension in the cable at its top end? (b) What is the tension in the cable at its bottom end? (c) What is the tension in the cable at its middle? 7. [25] A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40 and the coefficient of kinetic friction is 0.20. (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of 18.0 N, what i the magnitude of the friction force? (0 If the monkey applies a horizontal force of 18.0 N, what is the box's acceleration? Extra Credit 110] A chair of mass 10.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F-43.0 N that is directed at an angle of 40.0° below the horizontal and the chair slides along the floor. Use Newton's laws to calculate the normal force that the floor exerts on the chair.Explanation / Answer
Since you didn't mentioned which one is to be answered I am answering the first one
Ans 5:
a)
T - mg = 0 (since no net force acting)
T = 6*9.8 = 58.8 N
b)
we have upward acceleretion = a = 3m/s^2
T - mg = ma
T = m(g+a) = 6(9.8+3) = 76.8 N
c)
here a = -3m/s^2
T - mg = ma
T = 6*(9.8-3) = 40.8 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.