Question

A hockey puck is given an initial speed of 5.0 m/s. If the coefficient of kinetic friction between the puck and the ice is 0.05, how far does the puck slide before coming to rest? Solve this problem using conservation of energy.

Answer 1

Answer:

d = 25.51 m

Explanation:

the law of the conservation of energy says that:

$E_i - E_f = W_f$

where $E_i$ is the inicial energy, $E_f$ is the final energy and $W_f$ is the work of the friction.

so:

$E_i$ = $\frac{1}{2} MV^2$

$E_f = 0$

where M is the mass and V the velocity.

also,

$W_f = U_kNd$

where $U_k$ is the coefficient of kinetic frictio, N is the normal force and d is the distance.

therefore:

$\frac{1}{2}MV^2=U_kNd$

also, N is equal to the mass of the hockey puck multiplicated by the gravity.

replacing:

$\frac{1}{2}m(5)^2=(0.05)(m(9.8))(d)$

canceling the m:

$\frac{1}{2}5^2=0.05(9.8)(d)$

solving for d:

$d = \frac{\frac{1}{2}5^2 }{0.05(9.8)}$

d = 25.51 m

 

Answer 2

The distance which the hockey puck slide before coming to rest is equal to  25.51 meters.

Given the following data:

• Initial speed = 5.0 m/s

• Coefficient of kinetic friction = 0.05

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$.

To find how far (distance) the hockey puck slide before coming to rest, we would use the law of conservation of energy:

According to the law of conservation of energy:

$K.E_i - K.E_f = W_f$

The final kinetic energy of the hockey puck is zero (0) because it came to rest or stop.

$K.E_i - 0 = W_f\\\\K.E_i = W_f\\\\\frac{1}{2}mv_i^2 = umgd\\\\\frac{1}{2}v_i^2 = ugd\\\\v_i^2 = 2ugd\\\\d = \frac{v_i^2}{2ug}$

Substituting the given parameters into the formula, we have;

Distance, d = 25.51 meters

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