rce of 40 N is required to hold a spring that has been stretched from natural length of 10 cm to a length of 15 cm. How much work is done tretching the spring form 15 cm to 18 cm?

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A: Given: 110110 N force is required to stretch,The spring from it's natural length of 3838 cm to a…

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A: To evaluate using the trapezoidal rule with n=10.

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Q: Suppose that 4 J of work is needed to stretch a spring from its natural length of 32 cm to a length…

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Q: Suppose that 3 of work is needed to stretch a spring from its natural length of 30 cm to a length of…

A: we know the work needed to stretch the spring W=12kx2 where x is the elongation. x=xf-xi=45-30=15…

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Q: Suppose that 3 of work is needed to stretch a spring from its natural length of 30 cm to a length of…

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Q: A cable with mass 2 kg/meter is lifting a load of 150 kg that is initially at the bottom of a…

A: Mass = 150 kg, We know that, g = 9.81 ms2

Q: 2. Suppose that 2 J of work is needed to stretch a spring from its natural length of 30 cm to a…

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Q: What is the amount of work done on a spring when it is compressed from its natural length of 1 meter…

A: Compress from the natural length through 1-0.8= 0.2 meter spring constant 'k'=16 N/m

Q: Consider a spring with a natural length of 10.0 cm. It takes 4 Joules of work to stretch the spring…

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Q: Suppose that 3 ] of work is needed to stretch a spring from its natural length of 30 cm to a length…

A: Given problem:- Suppose that 3 of work is needed to stretch a spring from its natural length of 30…

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Q: A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a…

A: Law used:  Hooks Law: Fx=kx where k is the spring constant.

Q: A spring has a natural length of 10 cm. If it takes 5 J to stretch the spring from 10 cm to 15 cm.…

A: Consider the provided question, (16) Given, natural length of spring = 10 cm It takes 5 J to stretch…

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Q: If 6 J of work are needed to stretch a spring from 11 cm to 13 cm and 10 J are needed to stretch it…

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Q: A force of 6 pounds is required to hold a spring stretched 0.2 feet beyond its natural length. How…

A: F = -kx  6 = -k(0.2) So, k = 30   Work done, W = (1/2)kx2 = (1/2)*(30)*(0.9)2 = 12.15

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