Answer:
H=1020.12m
Explanation:
From a balance of energy:
[tex]\frac{m*Vo^2}{2} -mg*H=-Ff*d[/tex] where H is the height it reached, d is the distance it traveled along the ramp and Ff = μk*N.
The relation between H and d is given by:
H = d*sin(30) Replace this into our previous equation:
[tex]\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*N*d[/tex]
From a sum of forces:
N -mg*cos(30) = 0 => N = mg*cos(30) Replacing this:
[tex]\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*mg*cos(30)*d[/tex] Now we can solve for d:
d = 2040.23m
Thus H = 1020.12m
Answer:
the vertical height reach by the puck is 329.06m
Explanation:
In all the process, the only non-conservative force presented in the problem is the frictional force. Therefore, applying the Mechanical energy conservation:
[tex]\Delta E_{M} =W_{ncf}[/tex]
with:
[tex]E_{Mi}=\frac{1}{2} mv_{i} ^{2}\\E_{Mf}=mgH[/tex]
[tex]W_{ncf}=\int\limits^L_0 {\vec{F_{roz}} } \,\cdot \vec{dx}=-|F_{roz}|L=-\frac{H|F_{roz}|}{sin(\alpha)}[/tex]
From the dynamic analysis:
[tex]F_{roz}=\mu_{k}N=\mu_{k}cos(\alpha)mg[/tex]
Therefore:
[tex]E_{Mf}-E_{Mi}=W_{ncf}[/tex]
[tex]mgH-\frac{1}{2} mv_{i} ^{2}=-\frac{Hmg\mu_{k}}{tan(\alpha)}\\H-\frac{v_{i} ^{2}}{2g}=-\frac{H\mu_{k}}{tan(\alpha)}\\H(1+\frac{\mu_{k}}{tan(\alpha)})=\frac{v_{i} ^{2}}{2g}\\H=\frac{v_{i} ^{2}}{2g(1+\frac{\mu_{k}}{tan(\alpha)})}\\H=329.06m[/tex]
