Work, Energy & Velocity

Work, Energy & VelocityWorking Problems#1 A man exerted 140 N to move a cart at 1.05 m/s. He made a full stop when choosing an item to be bought in 7 seconds. How much work has done:(a) If the force exerted is parallel to the cart(b)If the force exerted is 300 with respect to horizontalSoln:F– 140 NVo – 1.05 m/sVf - 0 m/sT – 7 seca = (Vf – Vo) / t= (0 m/s – 1.05 m/s) / 7= - 0. 15 m/s2D = (Vf2 – Vo2) / 2a= [(0m/s)2 – (1.05 m/s)2 ]/ 2 ( - 0.15m/s2)= 3. 68 m(a) ParallelW = FD= (140 N) (3.68 m)Final answer => 515. 2 J(b) With ѲW = FcosѲD= (140 N)( cos300)(3.68 m)Final answer=> 446.18 J# 2A bus is moving at 60 km/hr has a momentum of 10,000 kg·m/s. What is its Kinetic energy?Soln:Step A. Convert: km/hr to m/s60 km/hr = 1000 m/ 1km x 1 hr/ 3600 s = 16.67 m/sB. Momentum = mass x velocity 10, 000 kg· m/s = mass (16.67 m/s) Mass = 599. 88 kgC. Plug inKE = 1/2mv2 = ½ (599.88kg)(16.67 m/s)2Final answer => 83,350 J# 3A 60- kg crate is acted upon by a horizontal force of 100 N. The object moves from rest and suddenly stops in 6 seconds. How much work has done: (a) If the floor is frictionless(b) If the coefficient of friction between the crate and the floor is 0.10Soln(a) ∑ Fx = ma 100 N = (60kg)(a) a = 1.67 m/s2D = Vot + (at2) / 2, but Vo = 0 So, = (at2) /2= [(1.67m/s2)(6s)2]/ 2= 30.06 mThen,W = FD = (100 N)(30.06 m) Final answer => 3,006 J (b)∑ Fx = maFx – fx = ma, if the object is acted with frictionSince, fx = FN So, Fx – FN = maFind FN:∑Fy = 0W – FN = 0 but, W = mgSo, mg – FN = 0 FN = mg = (60kg)(9.8m/s2) = 588 N Plug inFx – FN = ma100 N – [(588N)(0.10)] = (60kg)(a)100N – 58.8N = 60kg (a)41.2 N = 60kg (a) a = 0.69 m/s2then,D = Vot + (at2) / 2, but Vo = 0 So, = (at2) /2= [(0.69m/s2)(6s)2]/2= 12.42 m Eureka! W = FD = (100N)(12.42 m)Final answer => 1,242 J# 4An aircraft (F- 18) take off from USS Lincoln aircraft carrier moving straight North for a top classified air strike mission. The F- 18 aircraft takes 100 miles coasts due North then moves 120 miles, 300, East of North and finally moves 50 miles straight east. The aircraft will be expected to reach its respective target location in approximately 2 hours.(a) What is the velocity of the aircraft?(b) Suppose the enemy radar detected the F- 18 aircraft in Afghanistan. The enemy launched a guided missile which moves at constant speed of 60 m/s and can run out of fuel for about 25 minutes. If the aircraft moves at 40 m/s and its 35 km away from the place where the missile was lunched, Will the missile hit the aircraft? Note: Assume that the speed of the missile and the aircraft will remain constant.(a)Sketch N C B 300 A W E DR SA => 100 miles, north B => 120 miles, 300, East of NorthC => 50 miles eastD=> the shipNow,First, find Resultant Displacement (DR)1. DR -> F – 18The DR must be calculated from the point of origin ( the place where it starts to take off)Based on the sketch, the path of the aircraft forms a right triangleThus we assume that C in the Pythagorean Theorem is the DRDR = √∑X2 + ∑Y2But, first find the unknown x and y component of BBy studying this illustration: N C B ? 300 According to the Alternate Interior Angle Theorem of the Transversal, it follows that the unknown angle is also 300So, find the x and y components of B using the basic trigonometric functions (soh, cah , toa ,etc..)H as the hypotenuse which is 120 milesBx = cosѲH = cos300 (120 miles) = 103.92 milesBy = sinѲH = sin300 (120 miles) = 60 milesThen, proceed to the summation of the values∑X = 103.92 miles + 50 miles = 153.92 miles∑Y = 100 miles + 60 miles = 160 milesPlug in:DR = √ ∑X2 + ∑Y2 = √ (153.92 miles)2 + (160 miles)2 = 222.02 milesVoila!Velocity = Resultant displacement / time = 222.02 miles/ 2 hoursFinal answer => 111.01 miles/ hour(b)The enemy launched a guided missile which moves at 60 m/s and can run out of fuel for about 25 minutes. The aircraft moves at 40 m/s and its 35 km away from the place where the missile was lunched If we are going to analyze this statement, the enemy launched the missile when the aircraft is 35 km away from the place where the missile was launched. If the fuel of the missile will run out in 25 minutes, this means that the maximum distance that the missile can only travel must not be more than 25 minutes of time. In line with this, we also assume that the speed of the missile and the aircraft doesn’t change. Thus, in order to determine if the missile has the chance of hitting the aircraft is to find out the distance traveled by the aircraft in 25 minutes from the point that it is 35 km away from the opponent. Then, we must calculate the maximum distance that the missile can also travel in 25 minutes.The distance traveled by the aircraft in 25 minutes from the point that it is 35 km away from the opponentTD = D1 + D2D1 = 35 kmD2 = st = (40 m/s)(25 minutes x 60 seconds/ 1 minute) = 60,000 mTD = 95 kmThe maximum distance that the missile can only travel in 25 minutes.D=st = (60 m/s)(25 minutes x 60 seconds/ 1 minute) = 90 kmThe Final answerSo, Will the missile hit the aircraft? - NOWhy? The missile can only travel 90 km in 25 minutes which is shorter in order to reach the aircraft which can travel 95 km at the same time.