Why does this part not require a Sin(30) to decompose the MG(Sin(30))(.8)? The example gives a correct answer. 757.1 Gravitational Potential Energy215From Eq. (7.7),It would be very difficult to apply Newton's second law,EF = md, directly to this problem because the normal and frictionforces and the acceleration are continuously changing in both mag-nitude and direction as Throcky descends. The energy approach, bycontrast, relates the motions at the top and bottom of the ramp with-out involving the details of the motion in between.W; = Wother = K2 + Ugrav, 2 – Kj – Ugrav, 1= 450 J + 0 -0 - 735 J = -285 J%3DThe work done by the friction force is -285 J, and the totalmechanical energy decreases by 285 J.EVALUATE: Our result for W is negative. Can you see from thefree-body diagrams in Fig. 7.10 why this must be so?Example 7.6 An inclined plane with frictionWe want to slide a 12-kg crate up a 2.5-m-long ramp inclined at 30°. EXECUTE: (a) The energy quantities areA worker, ignoring friction, calculates that he can do this by giving itan initial speed of 5.0 m/s at the bottom and letting it go. But frictionis not negligible; the crate slides only 1.6 m up the ramp, stops, andslides back down (Fig. 7.11a). (a) Find the magnitude of the frictionforce acting on the crate, assuming that it is constant. (b) How fast isthe crate moving when it reaches the bottom of the ramp?K = }(12 kg)(5.0 m/s)² = 150 JUgrav, 1 = 0%3DUgrav, 2 = (12 kg)(9.8 m/s²)(0.80 m) = 94 J%3D%3DWather = - fsSOLUTIONHere s 1.6 m. Using Eq. (7.7), we findIDENTIFY and SET UP: The friction force does work on the crate asKi + Ugrav, 1 + Wother = K2 + Ugrav.2it slides. The first part of the motion is from point 1, at the bottomof the ramp, to point 2, where the crate stops instantaneously(v2 = 0). In the second part of the motion, the crate returns tothe bottom of the ramp, which we'll also call point 3 (Fig. 7.11a).We take the positive y-direction upward. We take y = 0 (andhence Ugrav(1.6 m)sin 30°%3DWother =-fs = (K2 + Ugrav, 2) - (K1 + Ugrav, 1)(++ + 4 J)- (150 J + 0) = -56 J = -fs%3D%3DWotherf :56 J= 35 N1.6 m= 0) to be at ground level (point 1), so that y1 = 0,0.80 m, and y3 =The friction force of 35 N, acting over 1.6 m, causes the mechani-Y2 =5.0 m/s. In part (a) our target variable is f, the magnitude of thefriction force as the crate slides up; as in Example 7.2, we'll findthis using the energy approach. In part (b) our target variable is v3,the crate's speed at the bottom of the ramp. We'll calculate thework done by friction as the crate slides back down, then use the0. We are given vi%3Denergy of the crate to decrease from 150 J to 94 J (Fig. 7.11b).(b) As the crate moves from point 2 to point 3, the work doneby friction has the same negative value as from point 1 to point 2.(The friction force and the displacement both reverse direction buthave the same magnitudes.) The total work done by frictionbetween points 1 and 3 is thereforecalenergy approach to find U3.Wother = Wfric = -2fs = -2(56 J) = -112 J7.11 (a) A crate slides partway up the ramp, stops, and slidesback down. (b) Energy bar graphs for points 1, 2, and 3.From part (a), K 150 J and Ugrav, 1gives= 0. Equation (7.7) then(a)The crate slidesupfrom point1 to point 2, then back downto its starting position(point 3).K + Ugrav, 1 + Wother = K3 + Ugrav, 3%3DPointK3 = K + Ugrav, 1 - Ugrav,3 + Wother150 J + 0 0+ (-112 J) = 38 J-2.5 m2U2 = 0-1.6 m%3DThe crate returns to the bottom of the ramp with only 38 J of the= 5.0 m/original 150 J of mechanical energy (Fig. 7.11b). Since K3mu%3DThe crate is movingat speed v, when itreturns to point 3.0.80 m30°2K32(38 J)U3 =2.5 m/s%3D12 kgPoint 1.3EVALUATE: Energy was lost due to friction, so the crate's speedU32.5 m/s when it returns to the bottom of the ramp is less thanthe speed v1The force of friction does negative work on5.0 m/s at which it left that point. In part (b) we1 and 3, considering the round trip as athe crate as it moves, so the total mechanicalenergy E = K + Ugravdecreases..Eq. (7.7) to po(b)whole. Alternatively, we could have considered the second part ofthe motion by itself and applied Eq. (7.7) to points 2 and 3. Try it;do you get the same result for v3?E = K+UgravE = K+UgravE = K+UrgravAt point 1At point (2At point 3zerozerozero

Question

Accepted Answer

In the given example, 
The problem based on the conservation of energy.