Q10E - University Physics with Modern Physics

Question

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. 

 

(a) What is the mass of the block of ice? 

 

(b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?

 

 

Step-by-Step Solution

Answer

 (a) The mass of the block of ice is $90.9 kg$ 

(b) If the worker stops pushing at the end of 5.00 s, the block moves $22 m$   in the next 5.00 S.

 

Step 1: Given data

The force applied to the block

  $F=80 \text{N}$

The distance traveled by the block while the force is being applied

  $S_1=11 \text{m}$

Time of application of force

  $t_1=5 \text{s}$

Time after which distance is to be calculated after the force is removed

 $t_2=5 \text{s}$

The initial velocity of the block is

  u=0

Step 2: Laws and equations of motion

The distance traveled s, time of travel t , 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 force F , 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 t  is the time.

Step 3: Mass of the block

Let the acceleration of the block be  . From equation (1),

    $\begin{array}{rcl}a& =& \frac{S_1-u{t}_1}{\frac{1}{2}{t}_1^2}\\ & =& \frac{11 \text{m}-0}{\frac{1}{2}{\left(5 \text{s}\right)}^2}\\ & =& 0.88 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

Thus from equation (2), the mass of the block is

  $\begin{array}{rcl}m& =& \frac{F}{a}\\ & =& \frac{80 \text{N}}{0.88 {\text{m/s}}^{\text{2}}}\\ & =& 90.9 \text{kg}\\ & & \end{array}$

The mass of the block is90.9kg.

Step 4: Distance traveled by block after force is removed

From equation (3), the final velocity of the block is

  $\begin{array}{rcl}v& =& u+a{t}_1\\ & =& 0+0.88 {\text{m/s}}^{\text{2}}\times 5 \text{s}\\ & =& 4.4 \text{m/s}\\ & & \end{array}$

From equation (4), the distance traveled in time   after the force is removed is

  $\begin{array}{rcl}{S}_2& =& v{t}_2\\ & =& 4.4 \text{m/s}\times 5 \text{s}\\ & =& 22 \text{m}\\ & & \end{array}$

Thus, the distance traveled is $22 \text{m}$ .