At a party, a 0.63 kg ball is going to be shot verticallyupward using a spring-l
ID: 1740733 • Letter: A
Question
At a party, a 0.63 kg ball is going to be shot verticallyupward using a spring-loaded mechanism. The spring has a forceconstant of 188 N/m and is initially compressed 45 cm fromequilibrium.How high will the ball go above the initial position, and what will its speed be at half of this height? (Assume the ball leaves contact with the spring when thespring reaches its usual equilibrium position.) At a party, a 0.63 kg ball is going to be shot verticallyupward using a spring-loaded mechanism. The spring has a forceconstant of 188 N/m and is initially compressed 45 cm fromequilibrium.
How high will the ball go above the initial position, and what will its speed be at half of this height? (Assume the ball leaves contact with the spring when thespring reaches its usual equilibrium position.)
Explanation / Answer
This is Hooke's law . Force provided by a spring is equal tothe spring constant multiplied by the distance from the spring'spoint of equilibrium, F = -kx. x is in meters. What you do with this problem is figure out the value of F for thespring, using the above formula F = -kx. From there, you setit equal to F = ma, and you have the acceleration of the ball asit's being launched by the spring. Use basic linear motionequations to determine the final velocity of the ball after it'smoved the 45 cm that the spring affects the ball. After you have the final velocity of the ball with regards to thespring accelerating it, you use that final velocity as the initialvelocity of a second linear motion problem, this time with gravity(-9.8) as your acceleration, and solve for the halfway point as youwould any other vertical launch problem (find the value for"t" where Vf = 0 m/s, that's the highest point of the ball. Determine the vertical height using those same basic linear motionformulas, divide that height in half, and finally solve for thevelocity at "h/2".)
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