The man slides the 100-kg crate across force of 200 N. If the crate was initially at rest, how far will the crate move before

its speed is 1 m/s? The coefficient of kinetic friction between the crate and the
floor is 0.18.

The man is pulling the crate at a 20 degree angle.

Well, it seems like the man is in quite a slippery situation, both literally and figuratively! Sliding a 100 kg crate with a 200 N force and a coefficient of kinetic friction of 0.18? That's definitely going to be a fun ride!

To start, let's break down the forces at play here. We have the force applied by the man, which is 200 N at a 20-degree angle. Then we have the force of kinetic friction opposing the motion, which we can calculate using the coefficient of kinetic friction and the normal force.

The normal force can be found using the equation: N = mg, where m is the mass of the crate (100 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, N = 100 kg * 9.8 m/s^2 = 980 N.

Now, we can calculate the force of kinetic friction using the equation: F_fric = μ * N, where μ is the coefficient of kinetic friction (0.18).

F_fric = 0.18 * 980 N = 176.4 N.

Since we're dealing with a friction force that opposes the motion, the net force acting on the crate is the difference between the applied force and the force of friction: F_net = F_applied - F_fric.

F_net = 200 N - 176.4 N = 23.6 N.

Now we can use Newton's second law, F = ma, to find the acceleration of the crate. Since the mass is 100 kg, we have:

23.6 N = 100 kg * a.

Solving for a, we get: a = 23.6 N / 100 kg = 0.236 m/s^2.

Next, we can use the kinematic equation, v_f^2 = v_i^2 + 2ad, where v_f is the final velocity, v_i is the initial velocity (0 m/s since the crate was initially at rest), a is the acceleration, and d is the displacement.

Plugging in the values, we have: (1 m/s)^2 = (0 m/s)^2 + 2 * 0.236 m/s^2 * d.

Simplifying, we get: 1 m^2/s^2 = 0 + 0.472 m^2/s^2 * d.

Now we can solve for d: d = 1 m^2/s^2 / 0.472 m^2/s^2 = 2.1186 m.

So, the crate will move approximately 2.1186 meters before its speed reaches 1 m/s. That's definitely a good distance to slide and test the limits of that coefficient of kinetic friction! Just make sure to have some popcorn and a good sense of humor ready for the occasion!

To find the distance the crate will move before its speed is 1 m/s, we need to analyze the forces acting on the crate.

1. Resolve the 200 N force into two components:
- The horizontal component: F_horizontal = F_pull * cos(angle)
- The vertical component: F_vertical = F_pull * sin(angle)

Given that F_pull = 200 N and the angle is 20 degrees:
F_horizontal = 200 N * cos(20 degrees)
F_vertical = 200 N * sin(20 degrees)

2. Calculate the frictional force:
The frictional force can be determined using the formula:
F_friction = coefficient_of_friction * Fn

The normal force (Fn) acting on the crate can be calculated using:
Fn = mass * gravity

Given that the mass of the crate is 100 kg and the acceleration due to gravity is approximately 9.8 m/s^2:
Fn = 100 kg * 9.8 m/s^2

Now we can calculate the frictional force:
F_friction = 0.18 * Fn

3. Determine the acceleration of the crate:
The net force acting on the crate is the difference between the horizontal component of the applied force and the frictional force:
F_net = F_horizontal - F_friction

Determine the acceleration using Newton's second law:
F_net = mass * acceleration
acceleration = F_net / mass

4. Calculate the time taken for the crate to reach a speed of 1 m/s:
The initial velocity of the crate is 0 m/s, and we want to find the time it takes to reach a velocity of 1 m/s.
The formula to find the time (t) is:
v_final = v_initial + acceleration * t

5. Calculate the distance travelled by the crate:
The distance traveled can be calculated using the equation:
s = v_initial * t + 0.5 * acceleration * t^2

Now let's plug in the values and calculate the distance the crate will move.

To find how far the crate will move before its speed is 1 m/s, we can use the equation of motion:

F_net = m * a

Here, F_net is the net force acting on the crate, m is the mass of the crate, and a is the acceleration.

The net force can be calculated as the difference between the applied force and the frictional force:

F_net = F_applied - F_friction

To find the applied force, we need to resolve the force into its horizontal and vertical components:

F_applied_horizontal = F_applied * cos(theta)
F_applied_vertical = F_applied * sin(theta)

Given that F_applied = 200 N and theta = 20 degrees, we can substitute these values:

F_applied_horizontal = 200 N * cos(20 degrees)
F_applied_vertical = 200 N * sin(20 degrees)

To calculate the frictional force, we use the formula:

F_friction = coefficient of kinetic friction * normal force

The normal force is equal to the weight of the crate, which can be calculated as:

normal force = m * g

Substituting the given mass of the crate (100 kg) and the acceleration due to gravity (9.8 m/s^2), we get:

normal force = 100 kg * 9.8 m/s^2

Now, substituting the given coefficient of kinetic friction (0.18) and the calculated normal force:

F_friction = 0.18 * normal force

Next, we can calculate the net force:

F_net = F_applied - F_friction

Finally, we use Newton's second law (F_net = m * a) to find the acceleration:

a = F_net / m

Once we have the acceleration, we can use the equation of motion to find the distance traveled by the crate before its speed is 1 m/s:

v^2 = u^2 + 2*a*s

Where v is the final velocity (1 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance traveled.

Rearranging the equation, we get:

s = (v^2 - u^2) / (2*a)

By plugging in the known values, you can calculate the distance traveled by the crate before its speed is 1 m/s.

net force=200cos20-(100*9.8-200sin20)*.18

but net force= 100*acceleration
solve for acceleration

then vf^2=vi^2+2ad solve for d