A 81 kg block is released at a 3.8 m height. the track is frictionless. the block travels down the track, hits a massless spring
1 answer:
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Answer:
F = 24 N
Explanation:
In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end
Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity
∑ τ = 0
F l -x W = 0
F =
let's calculate
F = 4/100 600
F = 24 N
Answer:
The speed is 6.1m/s with an angle of 56.3 degrees north of east.
Explanation:
We meed to apply the Linear momentum theorem:
The magnitude of the velocity is given by:
and the angle: