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What is the net force acting on a sliding crate?
The net force on a sliding crate is the vector sum of all forces acting on the crate. It is the force that is causing the crate to accelerate or decelerate. If all forces are balanced, the net force will be zero and the crate will move at a constant velocity.
How do you solve a 200-lb crate lies on the floor. The coefficient of static friction between the crate and the floor is 0.60. What is the minimum force required to start the crate sliding?
To calculate the minimum force required to start the crate sliding, you would multiply the weight of the crate by the coefficient of static friction. In this case, 200 lb crate * 0.60 static friction coefficient = 120 lb minimum force needed to start the crate sliding.
You are pushing a kg wooden crate across the floor. the force of sliding friction on the crate is 90 N. how much force must you exert on the crate to keep it moving with a constant velocity?
To keep the crate moving with constant velocity, the force you exert must balance the force of sliding friction. In this case, you must exert a force of 90 N in the opposite direction of the sliding friction, so the net force on the crate is zero and it remains in motion at a constant velocity.
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.
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 acting on a sliding crate?
The net force on a sliding crate is the vector sum of all forces acting on the crate. It is the force that is causing the crate to accelerate or decelerate. If all forces are balanced, the net force will be zero and the crate will move at a constant velocity.
How do you solve a 200-lb crate lies on the floor. The coefficient of static friction between the crate and the floor is 0.60. What is the minimum force required to start the crate sliding?
To calculate the minimum force required to start the crate sliding, you would multiply the weight of the crate by the coefficient of static friction. In this case, 200 lb crate * 0.60 static friction coefficient = 120 lb minimum force needed to start the crate sliding.
You are pushing a kg wooden crate across the floor. the force of sliding friction on the crate is 90 N. how much force must you exert on the crate to keep it moving with a constant velocity?
To keep the crate moving with constant velocity, the force you exert must balance the force of sliding friction. In this case, you must exert a force of 90 N in the opposite direction of the sliding friction, so the net force on the crate is zero and it remains in motion at a constant velocity.
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.
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.
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.
A force diagram for a crate is pushed across the floor?
The force diagram for a crate being pushed across the floor would typically show a horizontal force in the direction of the push, a normal force (perpendicular to the surface), and a frictional force opposing the motion. The sum of these forces determines the crate's acceleration or equilibrium.
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.
How much work is performed when a 60 kg crate is pushed 20 m with a force of 40 N?
400 J
How do you get a crate in RuneScape?
You mean the thing that gets pushed around varrock and Falador? that's part of a quest, called "Garden Of Tranquility."
Once the crate is sliding how hard do you push to keep is moving at constant valocity?
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.
What force is needed to slide a 250-kg crate across the floor at constant velocity if the coefficient of sliding friction between a crate and a horizontal floor is 0.25?
The force needed to slide the crate at constant velocity is equal in magnitude but opposite in direction to the force of friction. The force of friction can be calculated as the product of the coefficient of friction and the normal force acting on the crate (weight of the crate). Therefore, the force needed would be 250 kg * 9.8 m/s^2 * 0.25 = 612.5 N.
