A spring of spring constant k=200nm1 is slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. Calculate the work done by the extending force.
Answers
Answered by
8
K=200n/m
X1=3.0cm
X2=5.0cm
W=1/2kx^2
W1=1/2×200×3^2
W1=900j
Wtot=1/2kx^2
Wtot=1/2×200×5^2
Wtot=2500j
Workdone by extending force is
w=Wtot-W1
w=2500-900
w=1600
X1=3.0cm
X2=5.0cm
W=1/2kx^2
W1=1/2×200×3^2
W1=900j
Wtot=1/2kx^2
Wtot=1/2×200×5^2
Wtot=2500j
Workdone by extending force is
w=Wtot-W1
w=2500-900
w=1600
Answered by
2
Answer:
0.16J
Explanation
Because we know that the formula of elastic energy for a string is Ek=0.5kΔx² in this case, we have to find the value of both the extension for 3.0cm and 5.0cm.
an extension for 3.0cm: 0.5×200×3²=900J
an extension for 5.0cm: 0.5×200×5²=2500J
2500-900=1600
1600J is NOT the proper answer for this, the unit for the extension should be in METER, therefore
1600×0.0001=0.16J
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