A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 140 . How far does the crate travel in 10.0s? What is its speed at the end of 10.0s?

F=ma

a=F/m
a= 140Kg.m/s² / 32.5kg
a= 4.30769m/s²

d=Vot + 1/2at²
d=1/2at²
d=1/2(4.31m/s²)(10s)
d= 215.5m

Oh boy, we have a crate on the move! Let's calculate the distance it travels and its speed at the end of 10.0 seconds.

To find the distance traveled, we can use the equation: distance = (net force * time^2) / (2 * mass).

Plugging in the given values, we have:
distance = (140 N * (10.0 s)^2) / (2 * 32.5 kg).

Calculating this, we get the distance traveled by the crate. Give me a sec to crunch the numbers... Drums roll, please... The distance traveled by the crate is approximately 215.38 meters. That's quite a journey!

Now, let's move on to the speed at the end of 10.0 seconds. We can use the equation: speed = (net force * time) / mass.

Using the given values, we get:
speed = (140 N * 10.0 s) / 32.5 kg.

Calculating this, we find that the speed of the crate at the end of 10.0 seconds is approximately 43.08 m/s. That crate sure knows how to pick up the pace!

So, to summarize:
- The distance traveled by the crate in 10.0 seconds is approximately 215.38 meters.
- The speed of the crate at the end of 10.0 seconds is approximately 43.08 m/s.

Hope that answers your question in a fun and informative way!

To find the distance traveled and the speed of the crate, we can use the formulas of kinematics:

1. Calculate the acceleration of the crate using Newton's second law:

F = m * a
a = F / m
a = 140 N / 32.5 kg
a ≈ 4.31 m/s^2

2. Use the kinematic equation to find the distance traveled by the crate:

d = v0 * t + (1/2) * a * t^2
Since the crate is initially at rest, v0 = 0
d = (1/2) * a * t^2
d = (1/2) * 4.31 m/s^2 * (10.0s)^2
d ≈ 215.5 m

3. Use the kinematic equation to find the final velocity (speed) of the crate:

v = v0 + a * t
v = 0 + 4.31 m/s^2 * 10.0s
v ≈ 43.1 m/s

Therefore, the crate travels approximately 215.5 meters and has a speed of approximately 43.1 m/s at the end of 10.0 seconds.

To answer these questions, we need to use the concept of Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. From this, we can determine the acceleration of the crate.

1. Calculate the acceleration:
Net force (F) = 140 N
Mass (m) = 32.5 kg

Applying Newton's second law:
F = m * a

Rearranging the equation to solve for acceleration:
a = F / m
a = 140 N / 32.5 kg
a = 4.308 m/s^2

The crate's acceleration is 4.308 m/s^2.

2. Calculate the distance traveled:
To find the distance traveled by the crate, we can use the equation of motion:

d = v0 * t + (1/2) * a * t^2

where:
d = distance traveled
v0 = initial velocity (0 m/s, since the crate starts at rest)
t = time taken (10.0 s)
a = acceleration (4.308 m/s^2)

Substituting the values into the equation:
d = 0 * 10.0 + (1/2) * 4.308 * (10.0)^2
d = 0 + 0.5 * 4.308 * 100
d = 215.4 m

The crate travels a distance of 215.4 meters in 10.0 seconds.

3. Calculate the final velocity:
We can find the final velocity of the crate using the formula:

v = v0 + a * t

where:
v0 = initial velocity (0 m/s)
t = time taken (10.0 s)
a = acceleration (4.308 m/s^2)

Substituting the values into the equation:
v = 0 + 4.308 * 10.0
v = 43.08 m/s

The crate has a velocity of 43.08 m/s at the end of 10.0 seconds.

Therefore, the crate travels a distance of 215.4 meters and has a speed of 43.08 m/s at the end of 10.0 seconds.

F=ma,or acceleration=force/mass

distance=1/2 a*t^2

vf=a*t