Answer to Question #234657 in Mechanics | Relativity for Elam

Question

3. A particle of mass 6 kg is on a slope at an angel of 20o
to the horizontal. The coefficient of
friction between the particle and the slope is 0.1. The particle is 5 m from the bottom of the
slope. It is projected up the slope with a speed 4 m s-1
?
(a) Find the distance travelled up the slope from the starting point until the particle comes
to rest.
(b) Find the time until the particle reaches the bottom of the slope.

Answer 1

$\text{Let's denote the X-axis along the slope}$

$\text{}$$\text{Take the positive direction up to the top }$

$\text{of the inclined plane}$

$\text{We choose the Y- axis perpendicular to the X-axis}$

$\alpha=20\degree$

$m= 6kg$

$g= 9.8\frac{m}{s^2}$

$\mu = 0.1$

$v_0= 4\frac{m}{s}$

$h_0= 5m$

$(a)$

$\text{Projection of forces on an Y-axis:}$

$N-mg\cos \alpha= 0$

$N=mg\cos \alpha= 6*9.8*\cos20 \degree=55.25N$

$\text{Projection of forces on an X-axis:}$

$F_{fr}= \mu N= 55.25*0.1=5.53N$

$F= - F{fr}-mg\sin\alpha= -5.53-6*9.8*\sin20 \degree=-25.65N$

$F = ma$

$a=\frac{F}{m}=\frac{-25.65}{6}=-4.28\frac{m}{s^2}$

$v = v_0+at$

$v_1 =0\text{ the particle has stopped moving}$

$t = -\frac{v_0}{a}=\frac{4}{4.28}=0.93s$

$S = v_0t+\frac{at^2}{2}= 4*0.93-\frac{4.28*0.93^2}{2}=1.87m(1)$

$\text{Answer: }1.87m$

$(b)$

$\text{Projection of forces on an Y-axis:}$

$N-mg\cos \alpha= 0$

$N=mg\cos \alpha= 6*9.8*\cos20 \degree=55.25N$

$\text{Projection of forces on an X-axis:}$

$F_{fr}= \mu N= 55.25*0.1=5.53N$

$F= -F{fr}+mg\sin\alpha=- 5.53+6*9.8*\sin20 \degree=14.58N$

$F = ma$

$a=\frac{F}{m}=\frac{14.58}{6}=2.43\frac{m}{s^2}$

$v_0=0$

$S _1= \frac{at^2}{2}$

$t = \sqrt{\frac{2S_1}{a}}$

$S _1= S(\text{from (1))}+\frac{h_0}{\sin\alpha}=1.87+14.2=16.07m$

$t = \sqrt{\frac{2S_1}{a}}= \sqrt{\frac{2*16.07}{2.43}}=3.64s$

$\text{Answer: }3.64s$