Example 5.10 Acceleration down a hill • A toboggan loaded with vacationing students (total weight w) slides down a long, snow-covered slope. The hill slopes at a constant angle α, and the toboggan is so well waxed that there is virtually no friction. What is its acceleration? a = g sinα n = mg cosα.
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Apr 21, 2008 · This means that the net force equals the component of the weight parallel to the ramp, and Newton's 2 nd Law says: This means that any object, regardless of size or mass, will slide down a frictionless ramp with the same acceleration (a fraction of g that depends on the angle of the ramp).

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  • Suppose You Shove A Puck Up The Ramp With Velocity Of 5 M/s. How Far From The Initial Point Does The Puck Turns Around? Suppose you shove a puck up the ramp with velocity of 5 m/s. How far from the initial point does the puck turns around?
  • 19. A 61 kg skateboarder is moving down a ramp with a 7.0 m radius of curvature. At the bottom of this ramp he reaches a speed of 7.8 m/s. What upward force acts on the skateboarder at the bottom of the ramp? A. 7.0 B. 5.3 x102N C. 6.0 x102N D. 1.1 Page 4 of6 r = 7.0m

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A 75 kg box slides down a 25.0 degree ramp with an acceleration of 3.60 m/s^2. Find coefficient between the box and the ramp. What acceleration would a 175 kg box have on this ramp?

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Suppose You Shove A Puck Up The Ramp With Velocity Of 5 M/s. How Far From The Initial Point Does The Puck Turns Around? Suppose you shove a puck up the ramp with velocity of 5 m/s. How far from the initial point does the puck turns around?

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Sep 28, 2017 · If the ramp has uniform friction, or no friction, acceleration will be constant (possibly zero, but constant, in any case). If the ramp has spotty friction—slick spot here, rough spot there—then the acceleration will be uneven. Similarly, when pushing uphill, the acceleration will depend on the force applied.

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+10 pts. Answered. A block slides down a frictionless inclined ramp. If the ramp angle is 17.0° and its length is 30.0 m, find the speed of the block as it reaches the bottom of the ramp, assuming it started sliding from rest at the top.

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It's the height of the bottom of the ramp or the bottom of the ramp yet is 0.25 meters minus 1/2 g t squared. So you want to solve for T. So, um, we are going to add 1/2 GT squared to both sides, so 1/2 G T squared is equal to 0.25 meters now multiply both sides by two. It divide both sides by acceleration due to gravity and take a square root.

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Two frictionless blocks, having masses of 8.0 kilograms and 2.0 kilo- grams, rest on a horizontal surface. If a force applied to an 8.0 kilogram block gives it an acceleration of 5.0 m/s2, then the same force will give the 2.0 kilogram block an acceleration of (1) 1.2 m/s2 (2) 2.5 m/s2 (3) 10. m/s2 (4) 20. m/s2

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Question 422431: A 3-kg object is released from rest at a height of 5m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant k=100/Nm. The object slides down the ramp and into the spring, compressing is a distance x before coming to rest.

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Acceleration down a hill. Now we are going to look at the problem of an object sliding down a frictionless inclined plane. Suppose a toboggan loaded with vacationing students (total weight w) slides down a long, snow-covered slope. The hill slopes at a constant angle a...

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