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

Answer:

a) $x_{f} = 3.126\,m$, b) $v_{f} = 3.126\,\frac{m}{s}$, c) $x_{f} = 21.882\,m$, d) $v_{f} = 6.252\,\frac{m}{s}$

Explanation:

a) Position of the puck:

The acceleration experimented by the puck is:

$a_{puck} = \frac{0.25\,N}{0.160\,\frac{m}{s^{2}} }$

$a_{puck} = 1.563\,\frac{m}{s^{2}}$

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

$x_{f} = \frac{1}{2}\cdot (1.563\,\frac{m}{s^{2}} )\cdot (2\,s)^{2}$

$x_{f} = 3.126\,m$

b) Speed of the puck:

The speed of the puck is computed as follows:

$v_{f} = (1.563\,\frac{m}{s^{2}} )\cdot (2\,s)$

$v_{f} = 3.126\,\frac{m}{s}$

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:

$x_{o} = 3.126\,m + (3.126\,\frac{m}{s} )\cdot (5\,s-2\,s)$

$x_{o} = 12.504\,m$

Likewise, the initial speed is $3.126\,\frac{m}{s}$.

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

$x_{f} = 12.504\,m+(3.126\,\frac{m}{s} )\cdot (7\,s-5\,s)+\frac{1}{2}\cdot (1.563\,\frac{m}{s^{2}} )\cdot (7\,s-5\,s)^{2}$

$x_{f} = 21.882\,m$

d) The final speed of the puck is:

$v_{f} = 3.126\,\frac{m}{s} + (1.563\,\frac{m}{s^{2}})\cdot (7\,s-5\,s)$

$v_{f} = 6.252\,\frac{m}{s}$