A cable with 19.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after b
1 answer:
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Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by
where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is
Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,
Therefore, the additional work needed is
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Explanation:
It is given that, the position of a particle as as function of time t is given by :
Let v is the velocity of the particle. Velocity of an object is given by :
So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :
At t = 0,
Hence, this is the required solution.