<p><span style=\"font-size: small;\"><span style=\"font-family: times new roman,
ID: 1964049 • Letter: #
Question
<p><span><span>A mass of m=84.4<span>kg</span> is moving up a plane inclined at 45.9deg to the horizontal with an initial velocity of (-2.8)m/s. The surface of the plane has a coefficient of static friction of 0.33 <a title="TeX" href="https://smccd.mrooms.net/filter/tex/displaytex.php?%5Csmall%28%5Cmu_S%3D0.33%29" target="popup"><img class="texrender" title="small(mu_S=0.33)" src="https://smccd.mrooms.net/filter/tex/pix.php/568bfcad4efd7eaabfae98f7daafed12.png" alt="small(mu_S=0.33)" /></a> and a coefficient of kinetic friction of .21 . </span></span><span><span><a title="TeX" href="https://smccd.mrooms.net/filter/tex/displaytex.php?%5Csmall%28%5Cmu_K%3D0.21%29" target="popup"><img class="texrender" title="small(mu_K=0.21)" src="https://smccd.mrooms.net/filter/tex/pix.php/d48f47a23111d0c50522c6ed40d68faf.png" alt="small(mu_K=0.21)" /></a></span></span><span><span>. Gravity (g=9.8m/s^2) acts straight down. How far along the plane will the mass move before coming to a stop? ( what is the displacement of the mass) <span>(m, three significant figures, Take down the plane as positive)</span></span></span></p>Explanation / Answer
The force of friction =N, where is the coefficient of friction and N is the normal reaction
as the body is moving, =k where k is the coefficient of kinetic friction
N = mgcos
force of friction acting along the plane = kmgcos = 0.21 x 84.4 x 9.8 x cos45.9 = 120.88 N (+ve because the force of friction opposes the relative motion and the body is moving up the plane, which is negative direction)
force due to gravity along the plane = mgsin = 84.4 x 9.8 x sin 45.9 =593.98N
net force = 120.88+593.98 = 714.86N
when the block comes to a stop, the work done by net force on the block = the initial kinetic energy of the block
net force x displacement = 1/2mv2
714.86 x S = 1/2 x 84.4 x 2.82
S = 0.463m
It covers 0.463m up the incline before coming to a stop
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