Its been a great day of new. frictionless snow. Julie starts at the top of the 6

Question

Its been a great day of new. frictionless snow. Julie starts at the top of the 60 degree slope. At the bottom, a circular arc carries her through a 90 degree turn, and she then launches off a 3.0-m-high ramp. How far horizontally is her touchdown point from the end of the ramp? The spring in the figure has a spring constant of 1000N/m. It is compressed 15 cm, then launches a 200 g block. The horizontal surface is frictionless. but the block's coefficient of kinetic friction on the incline is 0.20. What distance d does the block sail through the air?

Explanation / Answer

Conservation of energy means KE1+PE1=KE2+PE2, where point 1 is at top of slope and point 2 is just as she leaves the ramp. KE1=0 so

KE2=PE1-PE2=m * g * (25m - 3m)=215.6*m

KE2=1/2 * m * v2^2, = 215.6m

v2 =20.77meters/sec.

. The velocity components are vx0=v2*cos(30) and vy0=v2*sin(30). To figure out how long she is in the air, we take the vertical direction

y=y0+vy0*t - 1/2 * g * t^2, where y=0 when she lands, y0 is 3 meters.

0= 3 + 20.77sin 30 + 0.5 * (-9.8)t^2

we Use the quadratic equation to get t=2.38 seconds.

now x=vx0*t = 20.77 cos 30 * (2.38)= 42.8 m

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