Question

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are μs = 0.40 and μk = 0.30, respectively. The puck's initial speed is 99 m/s. What vertical height does the puck reach above its starting point?

Answer 1

Answer:

H=1020.12m

Explanation:

From a balance of energy:

$\frac{m*Vo^2}{2} -mg*H=-Ff*d$   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:

$\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*N*d$

From a sum of forces:

N -mg*cos(30) = 0    =>  N = mg*cos(30)   Replacing this:

$\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*mg*cos(30)*d$   Now we can solve for d:

d = 2040.23m

Thus H = 1020.12m

Answer 2

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:

$\Delta E_{M} =W_{ncf}$

with:

$E_{Mi}=\frac{1}{2} mv_{i} ^{2}\\E_{Mf}=mgH$

$W_{ncf}=\int\limits^L_0 {\vec{F_{roz}} } \,\cdot \vec{dx}=-|F_{roz}|L=-\frac{H|F_{roz}|}{sin(\alpha)}$

From the dynamic analysis:

$F_{roz}=\mu_{k}N=\mu_{k}cos(\alpha)mg$

Therefore:

$E_{Mf}-E_{Mi}=W_{ncf}$

$mgH-\frac{1}{2} mv_{i} ^{2}=-\frac{Hmg\mu_{k}}{tan(\alpha)}\\H-\frac{v_{i} ^{2}}{2g}=-\frac{H\mu_{k}}{tan(\alpha)}\\H(1+\frac{\mu_{k}}{tan(\alpha)})=\frac{v_{i} ^{2}}{2g}\\H=\frac{v_{i} ^{2}}{2g(1+\frac{\mu_{k}}{tan(\alpha)})}\\H=329.06m$