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$1.50 skier and block

  • From Physics: General-Physics
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  • Due on Feb. 27, 2010
  • Asked on Feb. 27, 2010 at 06:47:21AM
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PIFF414
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Q:

1) The Earth's surface, where g= 9.80 m/s squared
An Olympic skier coasts down a slope with an angle of inclination of 38 degrees. A)Neglecting friction, how many forces are acting on the skier? B) What is the acceleration of the skier? C)If the skier has a speed of 5.2 m/s at the top of the slope, what is his speed when he reaches the bottom of the 36 m long slope? 

2)The Earth's surface, where g= 9.80 m/s sqaured .
A horizontal force of 41 N acting on a block on a frictionless, level surface produces an acceleration of 2.6 m/s squared. A second block, with a mass of 4.1 kg , is dropped onto the first. What is the magnitude of the acceleration of the combination of blocks if the same force continues to act?

 (Assume that the second block does not slide on the first block.)

 

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  • Posted on Feb. 27, 2010 at 07:41:15AM
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vb68
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A:
Preview: ... B) the parallel to plane (parralel to movement) component of the gravitational force is Gp =m*g*sin(slpha)  and must be equal to the inertil force m* ...
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  • Posted on Feb 27, 2010 at 12:42:32PM
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Preview: ... 8 degrees. So a skier = 9.8(sin[38]} = 9.8*.6156=6.03 m/sec 2 C.  speed at top of slope of skier is 5.2 m/s, find speed at bottom if the slope length he skis is 36m.  The kinematic equation used to solve this part is: V f2 = V i2 + 2ad where V f is the velocity at the end of the slope, V i is the velocity at the top of the slope a = skiers acceleration found in part B. d = the distance skied of 36m V f2 = V i2 + 2ad= 5.2 2 + 2(6.03)(36) = 27.04 + 434.16 = 461.2 V f =sqrt[461.2] = 21.48 m/sec 2)The Earth's surface, ...
The full tutorial is about 485 words long .