Part (b) Write an expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline.
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- A car of mass m = 1100 kg is traveling down a θ = 14 degree incline. When the car's speed is v0 = 13 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is μk = 0.45. Calculate the distance the car travels down the hill L in meters until it comes to a stop at the endThe figure shows a container of mass m1 = 4.9 kg connected to a block of mass m2 by a cord looped around a frictionless pulley. The cord and pulley have negligible mass. When the container is released from rest, it accelerates at 2.2 m/s? across the horizontal frictionless surface. What are (a) the tension in the cord and (b) mass m2? (al Number Units The absolute tolerance is ± 0.1. (b) Number Units This answer has no units * (degrees) Save for Later Attempts: 0 of 3 used Submit Answer m kg m/s m/s^2 N/m kg-m/s or N-s N/m^2 or Pa kg/m^3 m/s^3 timesA car is moving at 20 m/s along a horizontal road has its brakes suddenly applied and eventually comes to rest. What is the shortest distance in which it can be stopped if the friction coefficient between tires and road is 0.90? Assume that all four wheels brake identically. If the brakes don't lock the car stops via static friction.
- 1 V:IV H.W.docx H.W: Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between m, (5kg) and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn't stretch, As m (2Kg) move down, m,move to the right. m. د قوة الشد Tو تسارع النظام .a علما بأن البكرة مهملة الكتلة والاحتكاك؟ 出 11 V:IV H.W.docx H.W: Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between m, (5kg) and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn't stretch. As m; (2Kg) move down, m, move to the right. m. د قوة الشد Tو تسارع النظام .a علما بأن البكرة مهملة الكتلة والاحتكاك؟ 出 l20 Son la 1Question 9. Traveling at a speed of 16.1 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.590. What is the speed of the automobile after 1.21 s have elapsed? Ignore the effects of air resistance. Ans: 9.1 m/s
- Problem 8: A car of mass m = 1020 kg is traveling down a 0 = 12 degree incline. When the car's speed is vo = 11 m/s, a mechanical failure causes all four of its brakes to lock. The y coefficient of kinetic friction between the tires and road is u = 0.45.You are driving through the Ouachita Mountains at 12m/s. You see a maple-leaf oak (Quercus acerifolia) crossing the road and hit the brakes, 20m away. The coefficient of kinetic friction between your tires and the road is μ_k = 0.7. Will you strike this endangered broadleaf? Justify your answer with a calculation.Your final answer should be the distance you will travel before coming to a stop.A block with mass m starting from rest at point A is sliding down a rough incline with kinetic coefficient of friction µ. The incline has angle 0 with respect to the horizontal sur- face. As the block slides down for a distance O 1. KB = m gs cos 0 – µ mgs cos 0 8, it passes point B. O 2. KB = mgs sin 0 + µ m gs sin 0 O 3. KB = mgs cos0 – µm gs sin 0 m O 4. KB = mgs sin 0 + µ mgs cos 0 v =0 O 5. KB = mgs sin 0 – µm g s sin 0 m О6. Кв — тgs sin@ — дтдs cos® B O 7. KB = m gs cos 0 + µ mgs cos 0 O 8. KB = m gs cos 0 + µ mgs sin 0 Find the kinetic energy of the block as it passes B.
- A 4.00-kg block is sent up a ramp inclined at an angle 0 = 29.0° from the horizontal. It is given an initial velocity VO = 15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is k coefficient of static friction is μg = 0.60. = 0.40 and the What distance D along the ramp's surface does the block travel before it comes to a stop? D = m %A block of mass m =1 kg, slides down a rough incline with constant velocity. The coefficient of kinetic friction between the block and the incline is µr, and the incline makes an angle 0 = 30° horizontal. Take g = 10 m/s2. The coefficient of kinetic friction µ is then equal to: v= constant with the O 0.577 O 0.466 O 0.422 O 0.364An object of mass 3 kg moves towards the negative x axis with a constant speed of 8 m/s and the coefficient of kinetic friction of the object with the ground is 0.39, so the value of the net force on the box is: