Question

A- A spring, of mass 146 g, is extended a distance of 11.0 cm from its equilibrium position, pointed straight up in the air, and released so that it is launched straight upwards. If the spring has a spring constant k = 161 N/m, at what height above its initial position will the spring have an equal amount (magnitude) of kinetic and stored gravitational potential energy (relative to its initial height). After being launched, the spring is remains at its equilibrium position. Report your answer in m with the acceleration due to gravity being 9.8 m/s2

B- Two, elastic-like proteins (143 N/m, 102 N/m) are connected end-to-end in a chain and an atomic force microscope pulls on them end of the second one with a force of 271 nN. The end of the first protein is connected to a stationary cell that can be viewed as being effectively immobile. Determine, in nm, how far the second protein will be displaced from its equilibrium position when the cell is stationary and the proteins are stretched.

Answer 1

(1) w=, ke?..( kinetic elastic energy). (1) 2 where k=force constant. w=workdone. e=extension. 1 w= V = -mv2....(kinetic energy)....(2) 2 where m=mass velocity equate eqn(1) and (2) 2 -ke' make v2 the subject of the formula 11 ...(1) from question; e=11.0cm= nm 100 146 k=161Nm-1; m=146g= kg. 1000 mv? 2 ke? va m v2= 11 161. 100 ; simplify more futher 146 1000 va=13.34ms-1 now from law of conservation. 1 2m my?=mgh 2 12=2gh v2 h= 2g (1) h=(13.34) =0.68m...(ans). 2.(9.8)

-1 (2) from hookes law. f=ke f=271nN (nano newton) k_=143Nm (stationary).. e=0 (k)=102Nm "cxtcnds) changing k, from Nm 102 -nNnm (nano form) 1,000,000,000 from law above. F=ke. F 271 -=2.7x10'nm(approx Idp) k 102 to nNnm-1 e 1x109