180 cos 28 = 159 N horizontal pull
net horizontal force = 159 - 124 = 35 N
Now if the question is what is the acceleration then it is
a = F/m = 35/53 = 0.66 m/s^2
If the crate starts from rest, how fast will it be moving after the man has pulled it a distance of 2.70 m?
net horizontal force = 159 - 124 = 35 N
Now if the question is what is the acceleration then it is
a = F/m = 35/53 = 0.66 m/s^2
If the crate starts from rest, how fast will it be moving after the man has pulled it a distance of 2.70 m?
To solve this, we'll need to break it down into manageable steps. First, we'll need to find the net force acting on the crate. This can be calculated by subtracting the frictional force from the pulling force.
Given that the magnitude of the frictional force is 124.0 N and the pulling force is 180.0 N, we can subtract the two to get the net force:
Net force = 180.0 N - 124.0 N = 56.0 N
Next, we'll need to find the horizontal component of the pulling force. This can be calculated using trigonometry. Since the force is at an angle of 28.0° above the horizontal, we can find the horizontal component by multiplying the pulling force by the cosine of the angle:
Horizontal force = 180.0 N * cos(28.0°) = 157.4 N
Now that we have the net force and the horizontal component of the force, we can use Newton's second law of motion to find the acceleration of the crate. Newton's second law states that force equals mass times acceleration:
Net force = mass * acceleration
So, we can rearrange the equation to solve for acceleration:
Acceleration = Net force / mass = 56.0 N / 53.0 kg = 1.06 m/s^2
Finally, we can use the kinematic equation to find the final velocity of the crate. The kinematic equation relates final velocity, initial velocity, acceleration, and displacement:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * displacement
Since the crate starts from rest, the initial velocity is 0. And the displacement is given as 2.70 m. Plugging in these values, we can solve for the final velocity:
Final velocity^2 = 0 + 2 * 1.06 m/s^2 * 2.70 m
Final velocity^2 = 5.76 m^2/s^2
Final velocity = √(5.76 m^2/s^2) = 2.40 m/s (approximately)
So, after the man has pulled the crate a distance of 2.70 m, it will be moving at a speed of about 2.40 m/s.
And there you have it! The solution to our little physics riddle. Hope I was able to assist you with a touch of humor along the way!
First, let's calculate the work done by the man in pulling the crate using the formula:
Work = Force * Distance * cos(theta)
where:
- Force = 180.0 N (the force applied by the man)
- Distance = 2.70 m (the distance the man pulls the crate)
- theta = 28.0° (the angle above the horizontal)
Plugging in the values, we have:
Work = 180.0 N * 2.70 m * cos(28.0°)
Next, we need to calculate the work done by the frictional force. The work done by a force is given by the equation:
Work = Force * Distance * cos(theta)
where:
- Force = 124.0 N (the magnitude of the frictional force)
- Distance = 2.70 m (the distance the man pulls the crate)
- theta = 180° (since the frictional force acts in the opposite direction to the displacement)
Plugging in the values, we have:
Work = 124.0 N * 2.70 m * cos(180°)
Now, let's calculate the net work done on the crate. The net work is the difference between the work done by the man and the work done by the frictional force:
Net Work = Work (done by the man) - Work (done by the frictional force)
Finally, we can use the work-energy principle to find the speed of the crate. The work done on an object is equal to its change in kinetic energy:
Net Work = Change in kinetic energy
We can write this as:
(1/2) * mass * velocity^2 = Net Work
where:
- mass = 53.0 kg (mass of the crate)
- velocity = unknown (the speed of the crate)
Rearranging the equation, we have:
velocity = sqrt((2 * Net Work) / mass)
Now, substitute the calculated net work and the mass into the equation to find the velocity.