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Once the crate is sliding how hard do you push to keep is moving at constant valocity?
Wiki User
∙ 12y ago
You need to push as much as the force of friction is.
Wiki User
∙ 12y ago
You should push with a force equal to the force of friction acting on the crate. This will counteract the friction force and allow the crate to continue moving at a constant velocity. Pushing with a greater force will accelerate the crate, while pushing with a force lower than the frictional force will cause it to decelerate.
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Why is it more difficult to slide a crate starting from rest than it is to keep moving once it is sliding?
It is more difficult to slide a crate starting from rest because static friction exists between the crate and the surface, requiring a greater force to overcome. Once the crate is already sliding, kinetic friction is less than static friction, making it easier to keep moving with a lower force.
What is the net force on a crate sliding at an unchanging speed when pushed with a steady force of 75 N?
The net force on the crate sliding at a constant speed is zero. This is because the applied force of 75 N is balanced by the frictional force opposing the motion. As a result, the crate does not accelerate, and the net force is zero.
A 1500 N crate is being pushed across a level floor at a constant speed. What will be the acceleration of the crate?
The acceleration of the crate will be zero since it is moving at a constant speed. This means that the net force acting on the crate is zero, so the forces pushing it forward are balanced by the forces resisting its motion.
What is the force of friction acting on a crate that slides across the floor?
The force of friction acting on a crate sliding across the floor is equal in magnitude but opposite in direction to the force applied to move the crate. It depends on the coefficient of friction between the crate and the floor, as well as the weight of the crate.
Pull horizontally on a crate with a force of 140 N and it slides across the floor in a dynamic equilibrium How much friction is acting on the crate?
If the crate is in dynamic equilibrium, the frictional force acting on it is equal in magnitude but opposite in direction to the applied force. Therefore, the frictional force acting on the crate is also 140 N.
How does the path of the crate appear to somebody on the airplane and to someone on the ground when a crate is dropped from an airplane flying horizontally at constant speed?
To someone on the airplane, the crate would appear to fall straight down due to its initial horizontal velocity matching the airplane's speed. To someone on the ground, the crate would follow a parabolic path because of gravity acting on it vertically while it moves horizontally due to its initial velocity.
Pull horizontally on a crate with a force of 140 N and it slides across the floor in a dynamic equilibrium How much friction is acting on the crate?
If the crate is in dynamic equilibrium, the frictional force acting on it is equal in magnitude but opposite in direction to the applied force. Therefore, the frictional force acting on the crate is also 140 N.
A 1500 N crate is being pushed across a level floor at a constant speed. What will be the acceleration of the crate?
The acceleration of the crate will be zero since it is moving at a constant speed. This means that the net force acting on the crate is zero, so the forces pushing it forward are balanced by the forces resisting its motion.
If you push on a heavy crate to the right and it slides what is the direction of friction on the crate?
The direction of friction on the crate is opposite to the direction in which it is sliding. In this case, since you are pushing the crate to the right, the friction will act to the left in order to oppose the motion.
When crate training do you put the crate near you at night?
In my opinion you should start with the dog crate next to you, and start moving it away from you little by little.
What is the net force on a crate sliding at an unchanging speed when pushed with a steady force of 75 N?
The net force on the crate sliding at a constant speed is zero. This is because the applied force of 75 N is balanced by the frictional force opposing the motion. As a result, the crate does not accelerate, and the net force is zero.
If a crate weighing 508 N is resting on a plane inclined 26degrees above the horizontal if acceleration is 4.29 then how fast will the crate be moving after 6 seconds?
To determine the speed of the crate after 6 seconds, we first need to calculate the net force acting on the crate on the inclined plane. This can be done by resolving the weight of the crate into components parallel and perpendicular to the plane. Then, using Newton's second law, F = ma, where F is the net force, m is the mass of the crate, and a is the acceleration, we can find the acceleration down the incline. After finding this acceleration, we can use the kinematic equation v = u + at to calculate the final speed of the crate after 6 seconds, where v is the final velocity, u is the initial velocity (assumed to be 0), a is the acceleration, and t is the time.
What will happen at the exact moment a plane flies directly over a car the plane drops a crate where will the crate crash?
not on the car because they are not moving at the same speed
If a crate is stationary on an incline does it mean that it is in the state of equilibrium?
No, the crate being stationary on an incline does not necessarily mean it is in equilibrium. Equilibrium requires not only that the crate is stationary but also that the forces acting on it are balanced. Without knowing the exact forces acting on the crate, we cannot conclude that it is in equilibrium.
If a box is sliding along at a constant speed and the force of friction on the box is 100n what is the net force on the box?
If the box is sliding at a constant speed, the net force on the box is zero. The force of friction (100N) is balanced by an equal and opposite force exerted on the box to keep it moving at a constant speed.
A constant push of 250 N is needed to slide a crate that weighs 400 N along a 2 meter long ramp If the ramp raises the crate 1 meter what is the ramp's efficiency?
80%
A 40kg crate is at rest on a level surface if the coefficient of static friction between the crate and the surface is 0.69 what horizontal force is required to get the crate moving?
The force of static friction acts to prevent the crate from moving when at rest. To overcome this force and start the crate moving, a force larger than the force of static friction must be applied. The force required to get the crate moving can be calculated as the product of the coefficient of static friction and the weight of the crate, which is 40 kg * 9.8 m/s^2 = 392 N. So, the horizontal force required would be 0.69 * 392 N = 270.48 N.
