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Pre-Algebra
Equations, expressions, and inequalitie

maduregimc

Answered question

2021-12-16

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) What is the mass of the block of ice?
2) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?

Answer & Explanation

accimaroyalde

Beginner2021-12-17Added 29 answers

(1)
$x - x_{0} = 11.0 m$
$t = 5.00 s$
$v_{0 x} = 0$ .
$x - x_{0} = v_{0 x} t + \frac{1}{2} a_{x} t^{2} \Rightarrow$
PLAN:
the acceleration, then multiply by $\sum F_{x} = m a_{x}$
Let +x be the direction of the force.
$\sum F_{x} = 80.0 N$
$a_{x} = \frac{2 (x - x_{0})}{t^{2}} = \frac{2 (1.1 .0 m)}{{(500 s)}^{2}} = 0.880 \frac{m}{s^{2}}$
$m = \frac{\sum F_{x}}{a_{x}} = \frac{80.0 N}{0.880 \frac{m}{s^{2}}} = 90.9 k g$
(2) $a_{x} = 0$ and $v_{x}$ is constant.
After the first 5.0 s,
$v_{x} = v_{0 x} + a_{x} t = (0.880 \frac{m}{s^{2}}) (5.00 s) = 4.40 \frac{m}{s}$
The motion of the next $5 s (a = 0)$ will now move at this initial speed
$x - x_{0} = v_{0 x} t + \frac{1}{2} a_{x} t^{2} = (4.40 \frac{m}{s}) (5.00 s) = 22.0 m$
For (2),
first calculate the speed at the end of the period (5s) of applied force (accelerated kinematics)
Utilize this finishing velocity as the initial (and constant) value for the second component of the motion then.

Tiefdruckot

Beginner2021-12-18Added 46 answers

$x = x_{0} + v_{0 x} t + \frac{1}{2} a_{x} t^{2} = \frac{1}{2} a_{x} t^{2}$
or $a_{x} = \frac{2 x}{t^{2}} = \frac{2 (11 m)}{{(5 s)}^{2}} = 0.88 \frac{m}{s^{2}}$

nick1337

Expert2021-12-27Added 777 answers

Given
Force=80 N
Initial velocity $(u) = 0$
Distance moved 11 m in 5s
$s = u t + \frac{a t^{2}}{2}$
$11 = 0 + \frac{a \times 25}{2}$
$22 = a \times 25$
$a = \frac{22}{25}$
$a = 0.88 \frac{m}{s^{2}}$
and force = mass x acceleration
$80 = m \times 0.88$
$m = 90.9 k g$
Block acquires velocity at the conclusion of 5 seconds.
$v = u + a t$
$v = 0 + 088 \times 5 = 4.4 \frac{m}{s}$
Distance moved in next 5 sec
$s = v t + 0$
$s = 44 \times 5 = 22 m$

star233

Skilled2023-05-26Added 403 answers

Step 1) To find the mass of the block of ice, we can use Newton's second law of motion:

Force = mass \times acceleration

Since the block moves with a constant velocity, its acceleration is zero. Therefore, the applied force is equal to the force of inertia:

Force = mass \times acceleration = mass \times 0 = 0

Since the force applied by the dockworker is the only horizontal force acting on the block, the mass of the block of ice is 0 kg.
Step 2) If the worker stops pushing at the end of 5.00 s, the block will continue to move with its current velocity due to its inertia. Since there are no horizontal forces acting on the block, it will move with a constant velocity in the next 5.00 s.
Therefore, the block will move the same distance as it did in the first 5.00 s, which is 11.0 m.
Hence, the block will move 11.0 m in the next 5.00 s.

alenahelenash

Expert2023-05-26Added 556 answers

Answer:
1) The mass of the block of ice is approximately 90.9 kg.
2) If the worker stops pushing at the end of 5.00 s, the block will move approximately 11.0 m in the next 5.00 s.
Explanation:
To solve the given problem, let's use the following equations of motion:
1)

v = u + a t

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

v^{2} = u^{2} + 2 a s

Where:
-

s

represents the displacement (distance traveled)
-

u

represents the initial velocity
-

v

represents the final velocity
-

a

represents the acceleration
-

t

represents the time
Given:
The horizontal force applied,

F = 80.0

N
The displacement,

s = 11.0

m
The time,

t = 5.00

s
1) To find the mass of the block of ice, we'll use Newton's second law of motion,

F = m a

, where

m

is the mass and

a

is the acceleration. Since the frictional force is negligible, the only force acting on the block is the applied force. Therefore,

F = m a

.
We can rearrange the equation to solve for mass:

m = \frac{F}{a}

Now, let's calculate the acceleration using the given displacement and time.
Using equation 2,

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

, we can rearrange it to solve for acceleration:

a = \frac{2 (s - u t)}{t^{2}}

Substituting the given values into the equation, we get:

a = \frac{2 (11.0 - 0 \cdot 5.00)}{(5.00)^{2}}

Simplifying the equation:

a = \frac{22.0}{25.0}

Now, we can substitute the acceleration into the equation for mass:

m = \frac{80.0}{\frac{22.0}{25.0}}

Simplifying the equation:

m = \frac{80.0 \cdot 25.0}{22.0}

m \approx 90.9

kg
Therefore, the mass of the block of ice is approximately 90.9 kg.
2) If the worker stops pushing at the end of 5.00 s, the block will continue to move with the initial velocity it acquired during the first 5.00 s. So, the final velocity,

v

, remains the same.
To find the distance traveled in the next 5.00 s, we'll use equation 2:

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

Since the block starts from rest, the initial velocity,

u

, is 0. The time,

t

, is 5.00 s, and we already know the acceleration,

a

, from the previous calculation.
Substituting the values into the equation:

s = 0 \cdot 5.00 + \frac{1}{2} \cdot \frac{22.0}{25.0} \cdot (5.00)^{2}

Simplifying the equation:

s = \frac{22.0}{25.0} \cdot 12.5

s \approx 11.0

m
Therefore, the block will move approximately 11.0 m in the next 5.00 s.

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