A hockey puck struck by a hockey stick is given an initial speed v0 in the positive x-direction. The coefficient of kinetic friction between the ice and the puck is μk.

(a) Obtain an expression for the acceleration of the puck. (Use the following as necessary: μk and g.)
a =

(b) Use the result of part (a) to obtain an expression for the distance d the puck slides. The answer should be in terms of the variables v0, μk, and g only.
d =

(a) To obtain an expression for the acceleration of the puck, we need to consider the forces acting on it. The forces involved are the force applied by the hockey stick (in the positive x-direction), and the force of kinetic friction opposing the motion (in the negative x-direction).

We can start with Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Net force = ma

In this case, the net force is the force applied by the hockey stick minus the force of kinetic friction:

Net force = F_applied - F_friction

The force applied by the hockey stick is the only force causing acceleration, so we can represent it as:

F_applied = ma

The force of kinetic friction can be calculated using the coefficient of kinetic friction (μk) and the normal force (N). The normal force is equal to the weight of the puck (mg), where g is the acceleration due to gravity:

F_friction = μkN

Since the puck is on a horizontal surface, the normal force is equal to the weight of the puck:

N = mg

So, the force of kinetic friction becomes:

F_friction = μkmg

Substituting these values into the net force equation, we get:

ma = F_applied - F_friction
ma = ma - μkmg

Rearranging the equation to isolate the acceleration (a), we have:

a = (F_applied - μkmg) / m

Now, the puck is given an initial speed v0 in the positive x-direction. Initially, there is no friction acting on it, so the force applied is the only force acting on the puck:

F_applied = ma = m * a

Substituting this value back into the equation, we have:

a = (ma - μkmg) / m

Simplifying further:

a = a - μk * g

Therefore, the expression for the acceleration of the puck is:

a = -μk * g

(b) To obtain an expression for the distance d the puck slides, we can use the kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (which is zero in this case)
u = initial velocity (v0)
a = acceleration (-μk * g)
s = distance traveled (d)

Rearranging the equation to solve for d:

0 = v0^2 + 2(-μk * g) * d

Since the final velocity is zero, we get:

v0^2 = 2μk * g * d

Solving for d, we have:

d = v0^2 / (2μk * g)

Therefore, the expression for the distance d the puck slides is:

d = v0^2 / (2μk * g)

(a) The acceleration of the puck can be determined using the second law of motion:

Net force = mass * acceleration

The net force acting on the puck is the force of friction in the opposite direction of its motion:

Net force = - Frictional force

The frictional force is given by:

Frictional force = μk * Normal force

The normal force acting on the puck is equal to the weight of the puck:

Normal force = mass * gravity

Using these relationships, we can express the net force as:

Net force = - μk * mass * gravity

Since the only force acting on the puck in the x-direction is the frictional force, the net force is also equal to:

Net force = mass * acceleration

Setting these two expressions equal to each other, we have:

- μk * mass * gravity = mass * acceleration

Simplifying, we get:

a = -μk * g

Therefore, the expression for the acceleration of the puck is:

a = -μk * g

(b) To determine the distance the puck slides, we can use the equations of motion. The final velocity of the puck can be expressed as:

vf^2 = v0^2 + 2 * a * d

Since the puck comes to rest (vf = 0), we can solve for the distance d:

0 = v0^2 + 2 * (-μk * g) * d

Simplifying, we get:

d = v0^2 / (2 * μk * g)

Therefore, the expression for the distance the puck slides is:

d = v0^2 / (2 * μk * g)

a=Force/mass=mu*mg/m= mu*g in the direction opposite motion.

vf^2=vo^2+2ad
d= -vo^2/2a you finish it.