Consider a 100 kg box which is at the rest. Now, two forces are applied to the box and it starts to move. Consider F1 = 400 N and F2 = 500 N and e %3D 60. Here F2 is a horizontal force. Find (a) the %3D normal reaction force from ground (N), (b) the velocity if it slides 5 m. Take coefficient of kinetic friction to be 0.2. Assume g = 9.81 m/s². %3D
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- Three crates with masses m, = are connected on a rough floor with a coefficient of kinetic friction Hk 12 kg and m2 = m3 = 8 kg %3D %3D =0.15. Under the influence of an external force F, the three crates move to the right with a constant speed v = 1.2 m/s. What is the net force exerted on this system along the x axis, Fnet.x=? Motion my O Cannot be determined O 0.5 N O 0.15 N O 1.2 N4.56 CALC An object of mass m is at rest in equilibrium at the origin. At t = 0 a new force ... F (t) is applied that has components F. (t) = k1 + k2y F, (t) = kzt where k1, k2, and k3 are constants. Calculate the position r (t) and velocity v (t) vectors as functions of time.A 2-kg particle travels along a path defined byr= (3 + 2t2) m, θ = [(1/3)t3+ 2] rad. andz= (5 -2t2) m,where t is in seconds. Determine the r,θ, z components of force that the path exerts on the particleat the instant t= 1 s.
- The Omani "Noor Majan" car is of mass (M) kg is going up Jabal Akhdar, at angle (0), with (N) people inside the car each of (mp) kg. The car model provides (L) Newtons of lift, and it is accelerating up at (a) m / s2. The road provides a coefficienct of kinetic friction (uk). What is the force exerted by the engine? [Hint: This question has no numbers. Your answer needs to be an equation. ]A cart with mass 15.7 kg is intitally at rest. You get it moving by pushing on the cart at an angle of 0 = 37° as shown above. The magnitude of your force as a function of time is given by: Fyc(t) = Foe-bt, where b = 0.55 s-1 and Fo = 138.1 N. You can assume that the wheels roll perfectly on the ground. What is the speed of the cart when t = 3.3 s? Don't Know Where to Start? ... A Hint About N2L A Hint About the Proces ...... ........................ Vf = 10.25m/sA 45.0 kg box is pulled with a force of 205 N by a rope held at angle of 46.6˚ to the horizontal. The velocity of the box increases from 1.00 to 1.50 m/s in 2.50 s. Calculate:a) The net force acting horizontally on the box. b)The frictional force acting on the box. c) The horizontal component applied and d)The coefficient of kinetic friction between the box and the floor.
- A box of mass m = 25 kg slides with an initial speed v0 = 7.0m/s up a slope which forms the angle = 12° relative to the horizontal. A strong wind is blowing with fixed horizontals force Fv = 50N to the right (see figure). The coefficient of friction in motion is uk = 0.18. (a) What is the transverse force acting on the box from the slope? (b) How far, L, does the box slide up the slope before it stops? (c) What must be the minimum coefficient of friction at rest, us, for the box to slide not back down the hill? F₁ m Vo L DProblem 2: The 300 metric tonne jet airliner has four engines, each of which produces a nearly constant thrust of 180 kN during the takeoff roll. Determine the length s of runway required if the takeoff speed is 220 km/hr. Compute s first for an uphill takeoff direction from A to B and second for a downhill takeoff from B to A on the 0.5° Horizontal slightly inclined runway. Neglect air and rolling resistance. Ans. (a) S-807 m (b) sF751 mA young girl is riding a bicycle that has a total mass (including the kid) of 21.2 kg. The girl is moving at 5.99 m/s on a flat road when suddenly she slams on the brakes and skids to a stop in 8.22 meters. What was the magnitutde of the average force due to friction acting on the bicycle? Remeber, the magnitude of a force is always positive. Assume a constant friction force measured in Newtons and 3 significant digits in your answer.
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