Question

A dockworker applies a constant horizontal force of 73.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time 4.50 s.
(a)What is the mass of the block of ice?----- I found this to be 56.9kg and got it right.
(b)If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ? ------ I tried using [distance=0.5at^2] but it says its wrong. how do you do this question?

Answer 1

Concept and reason

The concepts that are to be used to solve the given problem are kinematic equations and net force.

First, find the acceleration of the block by using the kinematic equation that relates the initial speed, time, acceleration, and the distance travelled.

Then, by using the relation among the force, mass, acceleration, find the mass of the block.

Find the final speed achieved by the block by using the kinematic equation that relates the initial, final speed, time, and the acceleration.

Then, by using the relation among the speed, distance, and time, calculate the distance travelled with in the given time.

Fundamentals

The rate of change of distance is called speed and is given by,

$v = \frac{s}{t}$

Here, s is the distance and t is the time.

The equation of motion for distance is,

$s = ut + \frac{1}{2}a{t^2}$

Here, u is the initial velocity, a is the acceleration, and t is the time.

Newton’s second law of motion states that the net force acting on an object is directly proportional to the acceleration of the object and is given by,

$F = ma$

Here, m is the mass and a is the acceleration.

The kinematic equation that relates the final speed, initial speed, acceleration, and time is,

$v = u + at$

(a)

The equation of motion for the distance travelled is,

$s = ut + \frac{1}{2}a{t^2}$

The block starts from rest and hence the initial velocity is zero.

Substitute 0 for u and rearrange the equation for acceleration a.

$\begin{array}{c}\\s = \left( 0 \right)t + \frac{1}{2}a{t^2}\\\\ = \frac{1}{2}a{t^2}\\\\a = \frac{{2s}}{{{t^2}}}\\\end{array}$

Substitute $13.0{\rm{ m}}$ for s and $4.50{\rm{ s}}$ for t.

$\begin{array}{c}\\a = \frac{{2\left( {13.0{\rm{ m}}} \right)}}{{{{\left( {4.50{\rm{ s}}} \right)}^2}}}\\\\ = 1.28{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The force acting on the block is expressed as follows:

$F = ma$

Rearrange the equation for mass.

$m = \frac{F}{a}$

Substitute $73.0{\rm{ N}}$ for F and $1.28{\rm{ m/}}{{\rm{s}}^2}$ for a.

$\begin{array}{c}\\m = \frac{{73.0{\rm{ N}}}}{{1.28{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 56.855{\rm{ kg}}\\\end{array}$

Round off the value to three significant figures. So, the mass of the block is $56.9{\rm{ kg}}$ .

(b)

The final velocity is given by,

$v = u + at$

Here, u is the initial velocity, a is the acceleration, and t is the time.

The block is at rest initially. So the initial velocity is zero.

Substitute 0 for u, $1.28{\rm{ m/}}{{\rm{s}}^2}$ for a, and $4.50{\rm{ s}}$ for t.

$\begin{array}{c}\\v = 0 + \left( {1.28{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {4.50{\rm{ s}}} \right)\\\\ = 5.777{\rm{ m/s}}\\\end{array}$

The rate of change of displacement is called velocity and is given by,

$v = \frac{s}{t}$

Rearrange the equation for the displacement.

$s = vt$

Substitute $5.777{\rm{ m/s}}$ for v and $4.20{\rm{ s}}$ for t.

$\begin{array}{c}\\s = \left( {5.777{\rm{ m/s}}} \right)\left( {4.20{\rm{ s}}} \right)\\\\ = 24.26{\rm{ m}}\\\end{array}$

Round off the value to three significant figures. So, the distance moved by the block is $24.3{\rm{ m}}$ .

Ans: Part a

The mass of the block is $56.9{\rm{ kg}}$ .

Answer 2

SOLUTION ;

Ice block moves 13.0 m in 4.50 sec. Initial speed is zero.

Hence, using formula :

s = ut + 1/2 a t^2 

=> 13.0 = 0 + 1/2 a (4.5)^2

=> 13.0 = 10.125 a 

=> a = 13.0 / 10.125 = 1.284 m/sec^2 

a.

So,

Mass of the block, m 

= F/a 

= 73 / 1.284 kg 

= 56.86 kg 

Mass of ice block is = 56.86 kg (ANSWER).

b.

Speed of block at t = 4.50 sec :

v4.50 = u + at = 0 + 1.284 * 4.5 = 5.78 m/s

So distance moved in next 4.20 sec when force is no more applied : 

= v4.50  * (t) 

= 5.78 * 4.20

= 24.28 m 

Distance moved in next 4.20 sec = 24.28 m (ANSWER)