Q11E - University Physics with Modern Physics

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; she continues to apply this force until t = 2.00 s.

(a) What are the position and speed of the puck at t = 2.00 s?

(b) If the same force is again applied at t = 5.00 s, what are the position and speed of the puck at t = 7.00 s?

 

Step-by-Step Solution

Answer

(a) The position and speed of the puck at t = 2.00 s are 3.12 m and 3.12 m/s respectively.

(b) The position and speed of the puck at t = 7.00 s are 27.84 m and 6.24 m/s respectively.

Step 1: Given data

The mass of the puck is

  $m=0.16 \text{kg}$

The force applied on the puck from  $t=0 \text{s}$ to $t=2 \text{s}$  is

  $F=0.25 \text{N}$

The initial velocity of the block is

  $u = 0$

Step 2: Laws and equations of motion

The distance traveled , time of travel , initial velocity  u and acceleration   are related as 

     $S=ut+\frac{1}{2}a{t}^2 .....\left(1\right)$

The second law of motion relates the forceF , mass m  and acceleration   as

  $F=ma .....\left(2\right)$

The final velocity  ,  time of travel t , initial velocity   u and acceleration   are related as 

     $v=u+at .....\left(3\right)$

For uniform motion, the distance traveled is

  $S=vt .....\left(4\right)$

Here,v  is the velocity and   is the time.

Step 3: Position and velocity of puck at t=2s

Let the acceleration of the block be . From equation (2),

    $\begin{array}{rcl}a& =& \frac{F}{m}\\ & =& \frac{0.25 \text{N}}{0.16 \text{kg}}\\ & =& 1.56 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

From equation (1), the distance taveled by the puck during the initial application of force is

  $\begin{array}{rcl}S& =& ut+\frac{1}{2}a{t}^2\\ & =& 0+\frac{1}{2}\times 1.56 {\text{m/s}}^{\text{2}}\times 4 {\text{s}}^2\\ & =& 3.12 \text{m}\\ & & \end{array}$

From equation (3), the velocity of the puck after the initial application of force is

  $\begin{array}{rcl}v& =& u+at\\ & =& 0+1.56 {\text{m/s}}^{\text{2}}\times 2 \text{s}\\ & =& 3.12 \text{m/s}\\ & & \end{array}$

Thus, the position is 3.12m  and the velocity is3.12m/s.

Step 4: Position and velocity after the second application of force

Before the second force is applied, the puck moves with a constant speed for 3 s. From equation (4), distance traveled in this time is

  $\begin{array}{rcl}{S}_1& =& v\times 3 \text{s}\\ & =& 3.12 \text{m/s}\times 3 \text{s}\\ & =& 9.36 \text{m}\\ & & \end{array}$

From equation (1), distance traveled during the second application of force is

  $\begin{array}{rcl}{S}_2& =& v\times 2 \text{s}+\frac{1}{2}a\times {\left(2 \text{s}\right)}^2\\ & =& 3.12 \text{m/s}\times 2 s+\frac{1}{2}\times 1.56 {\text{m/s}}^{\text{2}}\times 4 {\text{s}}^2\\ & =& 9.36 \text{m}\\ & & \end{array}$

Thus, position after the second application of force is

  $$\begin{gathered} S+S_1+S_2 \\ =3.12 \text{m}+9.36 \text{m}+9.36 \text{m} \\ =27.84 \text{m} \end{gathered}$$

From equation (3), the velocity after the second application of force is

  $\begin{array}{rcl}{v}_2& =& v+at\\ & =& 3.12 \text{m/s}+1.56 {\text{m/s}}^{\text{2}}\times 2 \text{s}\\ & =& 6.24 \text{m/s}\\ & & \end{array}$

Thus, the position is $27.84 \text{m}$ and the velocity is  $6.24 \text{m/s}$