Question

A 1.0 x 10^2-kilogram box rests on the bed of a
truck that is accelerating at 2.0 meters per
second^2
.
What is the magnitude of the force of
friction on the box as it moves with the truck
without slipping?

1) 1.0 x 103

N 3) 5.0 x 102
N

2) 2.0 x 102

N 4) 0.0 N

Answer 1

Answer:

$2\times 10^2\ \text{N}$

Explanation:

$\mu$ = Coefficient of friction

g = Acceleration due to gravity

m = Mass of box = $1\times 10^2\ \text{kg}$

a = Acceleration of the block = $2\ \text{m/s}^2$

Acceleration is given by

$a=\mu g\\\Rightarrow \mu=\dfrac{a}{g}$

Friction force is given by

$f=\mu mg\\\Rightarrow f=\dfrac{a}{g}mg\\\Rightarrow f=ma\\\Rightarrow f=1\times 10^2\times 2\\\Rightarrow f=2\times 10^2\ \text{N}$

The magnitude of the force of friction on the box as it moves with the truck without slipping is $2\times 10^2\ \text{N}$.

Answer 2

The magnitude of the force of  friction on the box is equal to: 2. $2.0 \times 10^2\; Newton$.

Given the following data:

• Mass of box = $1.0 \times 10^2\;kilogram$

• Acceleration = 2 $m/s^2$

To determine the magnitude of the force of  friction on the box, we would apply Newton's Second Law of Motion:

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the force acting on the object and inversely proportional to its mass.

Mathematically, Newton's Second Law of Motion is given by this formula;

$Force = mass \times acceleration$

Substituting the given parameters into the formula, we have;

$Force = 1.0 \times 10^2 \times 2.0\\\\Force = 2.0 \times 10^2\; Newton$

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