Question

A hockey puck with mass 0.160 kg is at rest at the origin (x=0) on the horizontal, frictionless surface of the rink. At time t = 0 a player applies a force of 0.250 N to the puck, parallel to the x-axis; he continues to apply this force until t = 2.00s.

What is the position of the puck at t = 2.00s ?
In this case what is the speed of the puck?
If the same force is again applied at t = 5.00s , what is the position of the puck at t = 7.00s ?
In this case what is the speed of the puck?

Answer 1

a)
`x_(f) = 3.126\,m`, b)
`v_(f) = 3.126\,(m)/(s)`, c)
`x_(f) = 21.882\,m`, d)
`v_(f) = 6.252\,(m)/(s)`

Step-by-step explanation:

a) Position of the puck:

The acceleration experimented by the puck is:

`a_(puck) = (0.25\,N)/(0.160\,(m)/(s^(2)) )`

`a_(puck) = 1.563\,(m)/(s^(2))`

As the force remains constant during its time of application, accelaration is also constant. Position at given time is the following:

`x_(f) = (1)/(2)\cdot (1.563\,(m)/(s^(2)) )\cdot (2\,s)^(2)`

`x_(f) = 3.126\,m`

b) Speed of the puck:

The speed of the puck is computed as follows:

`v_(f) = (1.563\,(m)/(s^(2)) )\cdot (2\,s)`

`v_(f) = 3.126\,(m)/(s)`

c) The absence of external force and the fact that ground is frictionless lead to the conclusion that hockey puck moves out at constant speed from 2 s. to 5 s. Then, the initial speed is:

`x_(o) = 3.126\,m + (3.126\,(m)/(s) )\cdot (5\,s-2\,s)`

`x_(o) = 12.504\,m`

Likewise, the initial speed is
`3.126\,(m)/(s)`.

The new application of the same force means the return of a accelerated movement. Then:

`x_(f) = 12.504\,m+(3.126\,(m)/(s) )\cdot (7\,s-5\,s)+(1)/(2)\cdot (1.563\,(m)/(s^(2)) )\cdot (7\,s-5\,s)^(2)`

`x_(f) = 21.882\,m`

d) The final speed of the puck is:

`v_(f) = 3.126\,(m)/(s) + (1.563\,(m)/(s^(2)))\cdot (7\,s-5\,s)`

`v_(f) = 6.252\,(m)/(s)`