A mass m equals 4 kg on a slanted surface theta equals 30


A mass m equals 4 kg on a slanted surface theta equals 30 degrees. A light rigid cord runs from mass m to pully beta equals 1/2 and Mass of pully equals 2 kg which is free to rotate with negligible axial friction. Then the cord attaches to a larger mass 6 kg which is hanging vertically. The coefficient of kinetic friction between m and the incline plane is 0.200. The system is released from rest at t equals 0.

A) calculate the magnitude of the friction force on the 4 kg mass.

B) calculate the magnitude of the systems acceleration.

C) calculate the tension in the section of the cord attached to the smaller mass m.

D) calculate the tension in the section of the cord attached to the larger mass M.

E) given that the radius of the pully is 2 cm, calculate the angular acceleration of the pulley.

F) calculate the work done by friction on the system after the smaller mass has slid a distance of 2 meters up the plane.

G) calculate the change in potential energy of the entire system after the smaller mass m has slid a distance of 2 meters up the plane.

H) calculate the speed of the blocks after the smaller mass m has slid a distance of 2 meters up the plane.

Request for Solution File

Ask an Expert for Answer!!
Physics: A mass m equals 4 kg on a slanted surface theta equals 30
Reference No:- TGS0607050

Expected delivery within 24 Hours