A spring of stiffness 127 N/m, and with relaxed length 0.22 m, stands vertically

Question

A spring of stiffness 127 N/m, and with relaxed length 0.22 m, stands vertically on a table, as shown in the figure. Instead of compressing the spring with a heavy block, with your hand you push straight down on the spring until your hand is only 0.19 m above the table. (Assume that the positive y-axis points upward and is normal to the table.) What is now the vector L rightarrow , with the spring compressed? L rightarrow = m What is the magnitude of L? |L rightarrow | = m What is the unit vector L^? L^ = What is the stretch, s, including the correct sign? s = m What is the force F rightarrow exerted on your hand by the spring? F rightarrow = N

Explanation / Answer

the L is the vector of length for the spring. a. since the spring is standing vertically so the length vector will be in the direction of y axis and it is told in the question that when we push it the hand is 0.19 m from the table so the magnitude of L is 0.19 m if we consider j^ as unit vector in Y direction then L = 0.19 j^ b. magnitude = 0.19 m c. j is the unit vector in Y direction so L^ = j d. since we have pushed the spring below its natural length so stretch = 0.19-0.22 = -0.03 m e. force on hand = -k stretch = -127(-0.03 j^) = 1.38 j^ N hence force on hand is in Y direction with magnitude 1.38 N.

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