It goes across.
The force that resists the motion of the crate is the force of friction between the crate and the floor. This frictional force acts in the opposite direction to the pushing force applied by the person, making it harder to move the crate.
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.
The force of friction when a crate slides across the floor is dependent on the coefficient of friction between the crate and the floor, as well as the normal force acting on the crate. The frictional force resists the motion of the crate and can be calculated using the equation: frictional force = coefficient of friction * normal force.
The crate will move to the left because the force to the left (25N) is greater than the force to the right (10N). The net force acting on the crate is the difference between the two forces: 25N (left) - 10N (right) = 15N to the left.
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.
The force that resists the motion of the crate is the force of friction between the crate and the floor. This frictional force acts in the opposite direction to the pushing force applied by the person, making it harder to move the crate.
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.
The force of friction when a crate slides across the floor is dependent on the coefficient of friction between the crate and the floor, as well as the normal force acting on the crate. The frictional force resists the motion of the crate and can be calculated using the equation: frictional force = coefficient of friction * normal force.
The crate will move to the left because the force to the left (25N) is greater than the force to the right (10N). The net force acting on the crate is the difference between the two forces: 25N (left) - 10N (right) = 15N to the left.
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.
The magnitude of the force of friction on the crate is equal and opposite to the force you apply to push it, as long as the crate is moving at constant speed. This is because the force of friction balances the push force to prevent acceleration.
The magnitude of the force of friction on the crate would be equal to the magnitude of your push. This is because the crate is moving at a constant speed, indicating that the force you are applying is balanced by the force of friction acting in the opposite direction.
The first time you enter the room. Start by pushing the crate in the southeast corner across to the southwest. Then push the crate in the northwest corner across the room so it toches the iced crate. Stand behind the crate you just pushed and push it to the south. Then push the same crate westone space so it hits the first crate. Now push the crate one more time south and the switch will be activated. The second time you need to enter the room is different. start by pushing the crate to the north south. Go to the south and shove the crate to the middle of the puzzle so it passes over the middle floor switch. Now shove the crate that's on the eastern edge west then north. So Its sitting in front of the middle floor switch. Run to the crate that's furthest north and push it clockwise around the rim of the puzzle so it hits the crete you have yet to move. Finally face north and push the crate so it is on the switch leading upstairs.
It really depends. On the person and the type of crate. I think you need to be alot more specsific about that one im afraid. Also depends on the size of the crate. Type of animal..... You get the point. Sorry.
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.
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.
No, the work done on the crate by the rope is not zero. The work done is equal to the force exerted by the rope multiplied by the distance the crate is pulled. The fact that the crate pulls back on the rope in accordance with Newton's Third Law does not cancel out the work done by the rope.