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malfutka [58]
3 years ago
10

A wooden block with mass 1.05 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 deg

rees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.90m up the incline from A, the block is moving up the incline at a speed of 5.10 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is 0.55. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.
Physics
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

Explanation:

energy stored in spring initially

= kinetic + potential energy of block + energy dissipated by friction

= 1/2 mv² + mgh + μ mgcosθ x  d

m is mass , v is velocity at top position , h is vertical height , μ is coefficient of friction ,θ is angle of inclination of plane

= m (1/2 v² + gh + μ gcosθ x  d )

= 1.05 ( .5 x 5.1² + 9.8 x 4.9 sin35 + .55 x 9.8 cos35 x 4.9 )

= 1.05 ( 13.005 + 27.543 + 21.635)

= 65.3 J .

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A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
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W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

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7 0
3 years ago
Read 2 more answers
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