Find the constant (N/m) of a spring that is stretched 10.53 cm by a 641 g hanging object.

59.65 N/m

(or if we round and use 3 sig figs it is 60.0 N/m)

Explanation:

F=kx × F=mg

Step one:

we have to convert the centimeters to meters and the grams to kilograms. When converting cm to m you divide by 100. When converting g to kg, you divide by 1000. so:

  10.53÷100=0.1053 m    and  641÷1000= 0.641 kg

also since the problem mentions "hanging" we can assume this object is in earth and therefore this object has the acceleration of gravity. Which is 9.8m/s²

step 2:

time to plug in our variables, so,

( N/m x)(0.1053m)×(0.641 kg)(9.8m/s²)

step 3:

since we are missing a variable on the left side of this equation we have to multiply our variables on the right side and then work backwards.

so:(0.641 kg)(9.8m/s²)= 6.2818

and then we divide by our only given variable from the left (0.1053m) to both sides. and it'll end up looking something like this on your paper when showing work:

( N/m x)(0.1053m)×(0.641 kg)(9.8m/s²)

0.1053m                       0.1053m

so: 6.2818÷0.1053= 59.65 N/m or 60.0 N/m

hope this helped :)

the answer is b ) square root of 5