The spring has an unstretched length of 0.4m and a stiffness of 200N/m. The 3 kg slider and attached spring are released from rest at A and move in the vertical plane. (a) Compute the change in elastic potential energy from A to B. (b) Compute the change in gravitational potential energy from A to B. (c) Compute the velocity of the slider as it reaches B in the absence of friction.

The spring has an unstretched length of 0.4m and a stiffness of 200N/m. The 3 kg slider and attached spring are released from rest at A and move in the vertical plane. (a) Compute the change in elastic potential energy from A to B. (b) Compute the change in gravitational potential energy from A to B. (c) Compute the velocity of the slider as it reaches B in the absence of friction.

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter10: Virtual Work And Potential Energy

Section: Chapter Questions

Problem 10.57P: Find the stable equilibrium position of the system described in Prob. 10.56 if m = 2.06 kg.

See similar textbooks

Similar questions

2. Find the horizontal force P to start the motion of any part of the system of three blocks resting upon one another as shown. The weight of the blocks are W-300 KN, Ws= 100 KN and Wc =200 KN. Between A and B μ=0.30, between C and B, μ =0.20 and between C and the ground, p=0.10. 1 sni brod nog 02.0-4 B C 05.0-4 Solution: tamos quoy worla bre sosia asquar 005 300 en of jells400 N₁-300 F3 N-300- N₂=300 Assuming B and C will not move. F₁ = 300 (0.30) F₁ = 90 kN ne à bns A zelboß 1 P = 90 kN es ensla berb hebnu 02.0- y elin 05.0= q V in Celpris orllanimetsa MX 008 Assuming C will not move: al 1st F2=400 (0.20) = 80 kN OTE Bm ribirte F₂ vica Assuming the three blocks move: F3 = 0.10 (600) F3 = 60 kN P=60 KN Least force is 60 kN.
Th lamp , mass = 10 Kg, is suspended in the position shown. The undeformed length of spring AB is 0.84 m and the spring has a stiffness of K = 784 N/m. For L= 2.8 m find Tension in cable AC , Internal force in spring AB, length of AC
Those data are for ouestions 12.136 14. The slender 8-kg bar AB is horizontal and at rest. The spring has an unstretched length sọ of 1 mand a spring constant of 15 N/m. the length of the bar AB is l-1.5m - - Find the elastic energy due to the spring when the bar has rotate clockwise 0 - 45° after being released. Give your answer with 2 decimals.
the intermediate answers are given for the unit vectors for the cable tensions; however, all the work must be done to obtain these answers.
Solve the reaction forces.
Q4. (а) Explain the meaning of spring force and the graph of spring force against position
The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and cast is W=180 N. The horizontal distance between points A and B where the cables are attached to the leg is L=100 cm and the vertical distance is d=5 cm. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=30°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N) Answerb) Find the tensile force T2 in cable 2. (Write your result in N) Answerc) Find the angle α of cable 1 with the horizontal. Response
A cart is moving in a track which is shown in the bellow picture(both friction and rotation is not acting on this cart) v = 0 50 m 50 m 40 m 20 m (1)Draw an energy diagram as the cart moves on the track (kinetic energy, potential energy, and the total energy lines are required) (2)Label the points where the cart is in stable equilibrium, unstable equilibrium, and any turning points (3)Find the Maximum and Minimum speed.
Q3. As shown on the right in image below, the force Facting on the box varies with displacement s, and the coefficient in the image, C = 14. Determine the work done by force Fto the box when the box has displaced s = 1.9 m. Please pay attention: the numbers may change since they are randomized. Negative sign must be included if the work done is negative. Your answer must include 2 places after the decimal point, and proper SI unit. F (N) F 5 3 S 4 C s (m) 1 2
Two bodies is connected in point C, are supported by rigid underlay at an angle α. F is the force acting on right body. We are considering the weight of bodies G and the friction between the bodies and the underlay. Length of bodies l=2000mm.    G1= 160 N, G2= 180 N, α = 30° Find:  1.) Find the magnitude of the force so that it slips on the underlay 2.) Draw the dependence of the force F as a function of the coefficient of friction f (in the range 0-0.5)
The homogeneous box has mass m. The magnitude of force Pis slowly increased until motion occurs. Motion could occur as slipping or tipping and you must decide which occurs first. Find the value of P which first causes motion. P 30° Ho= 0.40 B 2.2 d Answer: P = i mg
Q2/ A belt drive consist of two V-belts in parallel, on grooved pulleys of the size. The angle of the groove is 30°. The cross – section area of each belt is 750 and u is 0.12. The density of the belt material is 1.2 Mg/m' and the maximun stress in the material is 7 MN/m?. Answer the following questions 1- The centrifugal tension in belts is (a) increases power transmitted (b) decreases power transmitted (c) have no effect on power transmitted (d) increases power transmitted up to a certain speed and then decreases