Question

In order to illustrate the Principle of Work and Energy, the dynamics professor stands near the top of the 30° inclined wedge. He then challenges students to remove a peg from the wedge that prevents block A of mass ?_A = 0.1 ?? from travelling up the plane and hitting him. The spring is firmly attached to both the fixed surface and to block A. The spring has an undeformed length of 1 ? but is initially compressed by 0.50 ?. The coefficient of kinetic friction between the block and wedge is ?_k = 0.87, the spring constant is ? = 8 N/m² , and the widths of the block and the professor can be assumed negligible compared to the other dimensions in the problem.

a) Determine the minimum distance ? that the professor can stand from the base of the wedge such that the block does not hit him.

b) Determine the distance ? that the disgruntled student tells the professor to stand from the base of the inclined wedge such that block ? hits him when its kinetic energy is maximum.

Answer 1

a)  The minimum distance ? that the professor stands from the base of the wedge such that the block does not hit him = 0.87 m

b)  Distance ? that the disgruntled student tells the professor to stand from the base of the inclined wedge such that block ? hits him when its kinetic energy is maximum = 0.662 m

The arrangenl of block and showing are : a of 0.5- Datum for Westinia a) Leto assume that q, is the distance by professor is blanding Velocity of block will become zero as it just reach. the professor The Principle of workle Energy states that, Work done by Al forces -. Change w Kinetic Energy Other form of this equation is, Worke done lay Eelaunal Fonce = ( Change in kinalis)if Change in ( Energy /'l Pelestial Energy). Went = AKE+APE 0 grital Velocity of Blecke A, ; - 0 Final Velocity of Block B, \ -o . Anelial Polenlial Energy of Block = {K(0.5)2. Final Polentol Enangy of Block = mgh. Change in Kinelic Energy - Final KE - Initial Ke (KE) of Change in KE , OKE = 0 Change in Polantial Energy (PE) - Final PE - Inlialpe APE = Magh- 2 K10.52

Friction force acting on Block A, f= MAN. where Nis Normal Reaction applied by wedge on block W = mg C6630 f=. 0-87x iguos 30 f = 0.739 N. Work done by friction will be negalue because it acts of hobile to the direction of dishlacement. Went = -0.739xX, = Force » Displacement Wert = -0.739% Subblilule the values in Equation , -8.739X = 0 + migh- 1 R(0.5) 2 - 0.739 X = 0·lX9-81 X2% – LX8X0.52 (1.962X) +(0.7398) -- 1 x8x0.52 x= 0.37m . Distance from base of wedge , X = % +05 X=0.37+005 a=0.87m k Minimum Distance & + Pole which block does not het professor b). The mus Kinetic energy of the block A will be maximum bolwen before the Shring is completely streeted stretched lie before! Sharing achieves its Natural lengen) Leto assume Bhat block moves me from initial position : Let us be the maxim a velocity at this froint,

Initial K.E=0. Final 'KE = 1 mu? ARE = I MU, 2 Initial PE = K (0-5) Final P.C. = 1kC0*5-x² + mgh Work done by external force, Wert = - ung cob30 x X, Wert = -0.739 X1 substitute the value in Equation (0, - 0 139 = 1 ml + J K (o^5-xi) + mah JK (65) I [h= 2x ] 4 mo.? = 4K60.5)2 - 00 730%1962%, - 14 K (0.5-2)=> Noue velocity will be mancimum at hoint where does = o. Differentiale Equalion 2 with respect Lm 20 de ..0 – 0.739 - 1,962-f4K*2(0-5-20)(–197 0 = -0.739 – 1962 + $ 2005-X) KX = (KxO05 ) - 0.739 – 1-962 21= (8x095) – 0.739–1962 dx to % 2= 0.162 m Distance of Professor from base, X = 2, +0.5 X=00162 +0.5 x= 0.662m k Location of Maximum Ringles Energy from the lease of Wedge