A crate of mass 100 kg rests on a horizontal floor.
A 100 kg crate sliding on the floor.
This is the maximum force that.
A f m 25n 100kg 0 25 m s 2.
A 100 kg crate sliding on a floor is brought to a stop by 25 n force.
One worker pushes forward on the crate with a force of 270 n while the other pulls in the same direction with a force of 130 n using a rope connected to the crate.
Here s how to crack it the normal force on the object balances its weight so f n mg 100 kg 10 m s 2 1 000 n therefore f static friction max f f static max µ s f n 0 4 1 000 n 400 n.
The maximum possible static friction force is µsn µsmg 0 4 100 10 400 n.
A uniform disc of mass and radius r is projected horizontally with velocity v c on a rough horizontal floor so that it starts off with a pure sliding motion at t 0.
The coefficient of static friction is 0 4.
Both forces are horizontal and the crate slides with a constant speed.
What is the deceleration of the crate.
After t 0 second it acquires pure rolling motion as shown in figure.
μ is the coefficeient of friction.
If a force of 250 n parallel to the floor is applied to the crate what s the magnitude of the force of static friction on the crate.
The coefficient of static friction between the crate and floor is 0 700 and the coefficient of kinetic friction is 0 600.
Both forces are horizontal and the crate slides at a constant speed.
One worker pushes forward on the crate with a force of 380 n while the other worker pull in the same direction with a force of 350 n using a rope.
Two workers are sliding a 300 kg crate across the floor.
What is the momentum of an 80 kg runner moving at a speed of 2 5 m s.
If a force of 250 n parallel to the floor is applied to this mass the magnitude of the force of static friction on the crate is 250 n.
Calculate the velocity of the centre of mass of the disc at t 0.
What is the crate s coefficient of kinetic friction on the floor.
A 20 0 kg crate is at rest on a floor as shown in.
Two workers are sliding a 100 kg crate across the floor.