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- Pp# 5: can you help me solve and understand this practice problem please(a-d)? A step by step explanation would be appreciated. Thank you!A block of mass m = 54 kg slides along a horizontal surface. The coefficient of friction between the block and the surface is μk = 0.37. The block has an initial speed of vo = 15 m/s in the positive x-direction as shown. Write an expression for the x-component of the frictional force the block experiences, Ff, in terms of the given variables and variables available in the palette. What is the magnitude of the frictional force in N? How far will the block travel, in meters, before coming to rest?It was a snow day, so Eric and some friends went sledding. He boarded his 60-inch sled and descended the steepest section near the tall trees. Before his ride ended at the location of a tree stump buried a few inches below the snow, he was accelerating at 2.41 m/s/s down the 20.1° incline. Determine the coefficient of friction between the sled and the snow. (Coefficients of friction are positive, so use the absolute value of your answer.) Use the approximation g ≈ 10 m/s2.
- help solve 4 part question ( iam aware its alot but really need help) Friction/Motion A) A 3 kg box sits on a table where the coefficient of friction is 0.4. Calculate the friction that will oppose any motion.B) A 1 kg box sits on a ramp with an angle of 7 degrees, where the coefficient of friction is 0.3. Calculate the friction that will oppose any motion. C) A 4 kg box sits on a table where the coefficient of friction is 0.3. A 29 N horizontal force pulls the box to the right. Find the acceleration. D) A 3 kg box sits on a ramp of 14 degrees where the coefficient of friction is 0.2. A 37 N force pulls the box uphill. Find the acceleration.The coefficient of static friction between a book and the level surface it slides on is 0.65. If the mass of the book is 2.0 kg, what minimum initial applied force is required to slide the books across the surface? Input your answer with correct significant digits and no units.a sled, with a mass=6.50kg, is sliding down a snow-covered hill that has a slope of 45.0 degrees. there is an applied force pushing down on the sled perpendicular to the incline of 80.4N. the coefficient of kinetic friction between the sled and the snow is .100. determine the acceleration of the sled down the slope. report your answer is meters per second squared, but only enter the numerical value.
- Consider the four factors listed below. For the real, physical experiment that was replicated using the M3 simulation, which of these factors would it have been necessary to ignore (i.e. assume to be negligible) in the test of Newton's 2nd Law that was done in Lab M3? Select all that apply. Options for selection: Friction in the pulley. Air resistance on the cart. Friction between the cart and track. Air resistance on the hanging-mass.The muzzle velocity of a typical 500 g spud is 25 m/s . The force given by the spud is K/(x+.09) where x is the distance of the barrel in meters. If the length of the barrel is 75 cm, what is the constant K? What is the force on the spud?Which of the following quantities can be described by their magnitude and direction?a. distanceb. energyc. forced. mass
- Monkey D. Luffy is sailing together with his crew, the “Straw Hat Pirates” in the West Philippine Sea. Thewind is quite strong and is exerting a 400N force on their sailboat “Thousand Sunny” directed to the south.The waves are also exerting a force of 208N westward. If Thousand Sunny has a mass of 300 kg (including thecrew and their luggages), what is the magnitude and direction of its acceleration? PLEASE ANSWER IMMEDIATELY AND SHOW FULL SOLUTION WITH FORMULAS AND UNITSTwo dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in (Figure 1) has a mass m=10kg and is being pulled by a force F on a table with coefficient of static friction μs=0.3. Four forces act on it: The applied force F (directed θ=30 above the horizontal). The force of gravity Fg=mg (directly down, where g=9.8m/s2). The normal force N (directly up). The force of static friction fsfs (directly left, opposing any potential motion). If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsNfs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: Fcosθ−μsN=0 Fsinθ+N−mg=0 In order to find the magnitude of force FF, we have to solve a system of two equations with both FF and the…What is the relationship between the accelerations experienced by each mass? (Write an equation showing how a1 and a2 are related?)