A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of friction of 2.7 N opposing the motion. what is the acceleration of the hockey puck? Choices:
0.0054 m/s^2
0.005 m/s^2
200 m/s^2
190 m/s^2

The magnitude of acceleration of the hockey puck is 190 m/s² .

Given data:

The mass of hockey puck is, m = 0.17 kg.

the magnitude of external force on the puck is, F = 35.0 N.

The magnitude of force of friction is, f = 2.7 N.

We need to find the net force acting on the puck first. For that consider the applied force in the x-direction. Then net force is,

F' = F - f

Solving as,

F' = 35.0 - 2.7

F' = 32.3 N

Now, use the Newton's Second law to obtain the magnitude of acceleration (a) as,

F' = ma

Solving as,

32.3 = 0.17 (a)

a = 32.3/0.17

a = 190 m/s²

Thus, we can conclude that the magnitude of acceleration of the hockey puck is 190 m/s² .

Learn more about the Newton's law of motion here:

link

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

So, the acceleration of the hockey puck is .

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.