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Due date: February 22, 2021
10:00 AM EST
5: Holt SF 05Rev 43 - 10.0 pts possible
A 0.290 kg block on a vertical spring with a
spring constant of 4.65 x 103 N/m is pushed
downward, compressing the spring 0.0500 m.
When released, the block leaves the spring
and travels upward vertically.
The acceleration of gravity is 9.81 m/s.
How high does it rise above the point of
release?
Answer in units of m.
x x

Respuesta :

Mass of block(m)=0.290 kg
Spring Constant k = 4.65×10³ N/m
Initial elongation =0.0500 m=5 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block


Pi = kx² / 2

Pi = (4.65×10³) (25×10−⁴) / 2

Pi = 5.81 J

Pf = m g h

Pf = (0.290) (9.81) (h)

Pi = Pf

5.81 = 0.29×9.81×h

h = 2.04 m

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