The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :
<em>PE</em> = <em>m g h</em>
<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)
<em>PE</em> = 58.8 J
Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :
<em>PE</em> = <em>KE</em>
58.8 J = 1/2 <em>m v</em> ²
117.6 J = (2.0 kg) <em>v</em> ²
<em>v</em> = √((117.6 J) / (2.0 kg))
<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s
Answer:
Explanation:
The given time is 1 / 4 of the time period
So Time period of oscillation.
= 4 x .4 =1.6 s
When the block reaches back its original position when it came in contact with the spring for the first time , the block and the spring will have maximum
velocity. After that spring starts unstretching , reducing its speed , so block loses contact as its velocity is not reduced .
So required velocity is the maximum velocity of the block while remaining in contact with the spring.
v ( max ) = w A = 1.32 m /s.
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