Homework Help: physics

Posted by joe on Sunday, November 6, 2011 at 2:41pm.

when a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm. A) what is the force constant of the spring. B) if the 2.50 kg object is removed how far will the spring strecth if a 1.25 kg block is hung on it. C) how much work must an external agent do to strecth the same spring 8.00cm from its unstreched position.

physics - Henry, Monday, November 7, 2011 at 8:57pm
Wo = mg = 2.50kg * 9.8N/kg = 24.5N. =
Weight of objedt.

A. Fk = Wo / d = 24.5 / 2.76 = 8.88N/cm

B. d=(1.25kg/2.50kg) * 2.76cm = 1.38cm

C. F = 8.88N/cm * 8cm = 71N.
W = Fd = 71 * 0.0888m = 6.31 Joules.
physics - Moses, Sunday, February 24, 2013 at 4:49am
In your "C" answer, second line, you should have used 0.08m, not 0.0888m (that was the K constant).

Therfore, the result is W=Fd= 71*0.08= 5.68 Joules

Answer this Question

Related Questions

physics - When a 2.50-kg object is hung vertically on a certain light spring ...
Physics - When a 3.00-kg object is hung vertically on a certain light spring ...
physics - When a 2.60-kg object is hung vertically on a certain light spring ...
physics - When a 4.48 kg mass is hung vertically on a certain light spring that ...
science(physics) - a object potential energy is described as 1/2kx^2. what does ...
physic - Hooke's law describes a certain light spring of unstretched length ...
PHYSICS!!! - Hooke's law describes a certain light spring of unstressed ...
PHYSICS - A spring is hung from a ceiling, and an object attached to its lower ...
physics - A light spring of constant k = 170 N/m rests vertically on the bottom ...
Physics - When a 0.50 kg-object is attached to a vertically supported spring, it...

For Further Reading

First Name:
School Subject:
Answer: