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Amani Amani
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2 years ago
In the figure, two identical springs have unstretched lengths of 0.25 m and spring constants of 300 N/m. The springs are attached to a small cube and stretched to a length L of 0.36 m as in Figure A. An external force P pulls the cube a distance D = 0.020 m to the right and holds it there. (See Figure B.) The work done by the external force P in pulling the cube 0.020 m is closest to




▸ 12 J.

▸ 0.80 J.

▸ 0.060 J.

▸ 6.0 J.

▸ 0.12 J.
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Essential University Physics

Essential University Physics


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danielaremu1danielaremu1
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2 years ago Edited: 2 years ago, xwykrhm
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#3
2 years ago
I had a similar question with different numbers, here's how I solved it:

In the figure, two identical ideal massless springs have unstretched lengths of 0.25 m and spring constants of 700 N/m. The springs are attached to a small cube and stretched to a length L of 0.30 m as in Figure A. An external force P pulls the cube a distance D = 0.020 m to the right and holds it there. (See Figure B.) The external force P, that holds the cube in place in Figure B, is closest to

Given Hooke's Law, F=kx where F is spring force, k is the constant and x is the position.

with changing position this will be delta x.

And F on the left side is equal to F on the right side

Using the formula above, the right side is F=(700)(0.3+0.02-.25)=49N

For the right side, F=(700)(0.3-0.02-0.25)=21N

Since the force from the left side needs to equal the force from the right, find the difference between these two forces to find the outside force P: (49N-21N)=28N

So the answer is 28N
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