Spring A has a force constant of 1.0N/m. Spring B has a force constant of 2.0N/m

Question

Spring A has a force constant of 1.0N/m. Spring B has a force constant of 2.0N/m. The force exerted by a spring is the product of the spring constant and the elongation of the spring. I will pull on spring A with Force A and I will pull on spring B with Force B.

Spring A is elongated by 5m in the direction that goes from ( 2, 3 ) to ( 5, 7 ). Spring B is elongated by 5m in the direction the goes from ( 2, 3 ) to ( 6, 0 ).

What is the magnitude of the sum of the two forces ( Force A added to Force B )? Give your answer in N to the tenth of a N but do not type the units

Explanation / Answer

FA = KA*xA

FA = 1*5

= 5 N

direction of F1, = (5-2)i + (7-3)j

= 3i + 4j

theta = tan^-1(4/3) = 53.13 degrees

FB = KB*xB

FB = 2*5

= 10 N

direction of F2, = (6-2)i + (0-3)j

= 4 i - 3j

theta= tan^-1(-3/4) = -36.9 degrees below +x axis

so, Fnet = FA + FB

|Fnet| = sqrt(FA^2 + FB^2 + 2*FA*FB*cos(53.13+36.7))

= sqrt(5^2 + 10^2 + 2*5*10*cos(53.13+36.7))

= 11.23 N <<<<<<<<------------Answer

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