We know that:
W=Fs
200J=150N*s
s=200J/150N
s=1,33m
The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N). Force is needed to stretch spring is 10.2 N.
<h3>What is spring force?</h3>
The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is
F = kx
The given data in the problem is;
F is the spring force =?
K is the spring constant= 8.5 N/m
x is the length by which spring got stretched = 1.2m
Hence the force is needed to stretch the spring is 10.2 N.
To learn more about the spring force refer to the link;
brainly.com/question/4291098
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Explanation:
Below is an attachment containing the solution.
True, scientists often talk to each other to figure out if their results were similar and what they could have done better.
Although, talking to other scientists does have risks, other scientists could copy your work and further better it.
So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol
