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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.

  1. Find the total mass occupied by the block of ice.
  2. If the worker quits moving at the end of 5s, how long does the block move in the next 5s?

This problem aims to familiarize us with the applied force and the acceleration of a moving body. The concepts required to solve this problem are from basic applied physics which include the sum of applied force, instantaneous speed, and newtons law of motion.

Let’s first look at instantaneous velocity, which notifies us how fast an object is moving at a particular instance of time, simply named velocity. It is basically the average velocity between two points. The only difference lies in the limit that the time between the two circumstances closes to zero.

\[ \vec{v} = \dfrac{x(t_2) – x(t_1)}{t_2 – t_1} \]

Expert Answer

We are given the following information:

A horizontal force $F_x = 80.0 \space N$,

The distance the car travels from rest $s = x – x_0 = 11.0 \space m$,

Part a:

First, we are going to find the acceleration using the newton equation of motion:

\[ s  = v_it + \dfrac{a_x t^2}{2} \]

Since the car starts from rest, so $v_i = 0$:

\[ 11  = 0 + \dfrac{a_x \times 25}{2} \]

\[ 22  = a_x\times 25 \]

\[ a_x = \dfrac{22}{25} \]

\[ a_x = 0.88 m/s^2 \]

Using the first equation of motion, we can find the mass of the object moving with an acceleration of $a = 0.88 m/s^2$:

\[ F_x = ma_x \]

\[ m = \dfrac{F_x}{a_x} \]

\[ m = \dfrac{80.0 N}{0.880 m/s^2} \]

\[ m = 90.9 \space kg \]

Part b:

At the end of $5.00 s$, the worker stops pushing the block of ice, which means its velocity remains constant as the force becomes zero. We can find that velocity using:

\[ v_x = a_x \times t \]

\[ v_x = (0.88 m/s^2)(5.00 s) \]

\[ v_x=4.4 m/s\]

So, after $5.00 s$, the block of ice moves with a constant velocity of $v_x = 4.4 m/s$.

Now to find the distance the block covers, we can use the distance formula:

\[ s=v_x\times t\]

\[ s=(4.4 m/s)(5.00 s)\]

\[s=22\space m\]

Numerical Result

The mass of the block of ice is: $m = 90.9\space kg$.

The distance the block covers is $s = 22\space m$.

Example

A worker drives a box with $12.3 kg$ on a horizontal surface of $3.10 m/s$. The coef­ficients of kinetic and static friction are $0.280$ and $0.480$, respectively. What force must the worker use to sustain the motion of the box?

Let’s set the coordinate so that the motion is in the direction of the $x$-axis. Thus Newton’s second law in scalar form appears like this:

\[F-f=0\]

\[N-mg=0\]

We know that frictional force $f=\mu k\space N$, we will get $f=\mu kmg$. Since the body is moving, we use the coefficient of kinetic friction $\mu k$.

Then we can rewrite the equation as:

\[F-\mu kmg=0\]

Solving for force:

\[F=\mu kmg\]

Substituting the values:

\[F=0.280\times 12.3\times 9.8\]

\[F=33.8\space N\]

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