a box of mass 5.7 kg slides across a floor and comes to a complete stop. if its initial speed was 10km/h and mu = 0.34 find the acceleration of the box

Question

An illustration of a physics experiment. A mid-sized brown cardboard box with a weight label marking it as 5.7 kg. This box is on a plain flat surface, perhaps a wooden floor with visible texture. The box is captured in motion as it slides with speed. Despite its momentum, the box's motion is gradually coming to a full stop due to the friction between it and the surface, visually marked by lines indicating slowing speed. Show a physics equation nearby for context, but without actual numbers or text. The lighting is bright and even.

a box of mass 5.7 kg slides across a floor and comes to a complete stop. if its initial speed was 10km/h and mu = 0.34 find the acceleration of the box

Answer 1

F=ma

-mu*mg=ma
solve for a

Answer 2

-Ffr=ma (because it is decelarating, therefore a=negative)

-19=5.7a
-19/5.7=a
a= -3.33m/s2

Answer 3

Well, well, well, it seems we have a sliding box conundrum! Let's see if we can use our clown physics to solve this one.

First, let's convert that initial speed from km/h to m/s because science doesn't like to have fun with units. So, 10 km/h is approximately 2.78 m/s.

Now, we can use the equation of motion for a sliding object:

final velocity squared = initial velocity squared + 2 * acceleration * distance

Since the box comes to a complete stop, its final velocity is 0 m/s. We also know the initial velocity is 2.78 m/s.

Now, if we assume that "mu" is referring to the coefficient of friction, we can use it to find the deceleration (which is acceleration in the opposite direction).

The equation for frictional force is: frictional force = coefficient of friction * normal force

But since the box is on a horizontal surface, the normal force is equal to the weight of the box which is the mass times gravitational acceleration (9.8 m/s^2).

So, the frictional force is equal to mu * m * g, where g is the acceleration due to gravity.

Using Newton's second law (F = ma), we can set the frictional force equal to the mass times the acceleration (ma).

mu * m * g = m * a

Canceling out the mass (I knew it would come in handy), we get:

mu * g = a

Now we just need to substitute the value of mu (0.34) and the acceleration due to gravity (9.8 m/s^2) into the equation and calculate:

a = 0.34 * 9.8 ≈ 3.33 m/s^2

So, the acceleration of the box is approximately 3.33 m/s^2.

Answer 4

To find the acceleration of the box, we can use the equation:

Acceleration (a) = Frictional force (Ff) / Mass (m)

To calculate the frictional force, we can use the equation:

Frictional force (Ff) = Coefficient of Friction (μ) * Normal Force (Fn)

The normal force (Fn) is the force exerted on the box perpendicular to the surface of contact. In this case, since the box is on a horizontal surface, the normal force is equal to the weight of the box, which can be calculated as:

Normal Force (Fn) = Mass (m) * Acceleration due to gravity (g)

where the acceleration due to gravity (g) is approximately 9.8 m/s^2.

Let's plug in the given values into the equations:

Mass (m) = 5.7 kg
Initial speed (v) = 10 km/h = 10,000 m / 3600 s = 2.78 m/s (converted to m/s)
Coefficient of friction (μ) = 0.34
Acceleration due to gravity (g) = 9.8 m/s^2

Now we can calculate the normal force:

Fn = m * g
Fn = 5.7 kg * 9.8 m/s^2
Fn = 55.86 N

Next, we can calculate the frictional force:

Ff = μ * Fn
Ff = 0.34 * 55.86 N
Ff = 18.99 N

Finally, we can calculate the acceleration:

a = Ff / m
a = 18.99 N / 5.7 kg
a ≈ 3.33 m/s^2

Therefore, the acceleration of the box is approximately 3.33 m/s^2.