Question

box. Mary applies a force of 72 N to push a box with an acceleration of 0.46 m/s?. When she increases the pushing force to 79 N, the box's acceleration changes to 0.64 m/s2. There is a constant friction force present between the floor and the (a) What is the mass of the box in kilograms? kg (b) What is the coefficient of kinetic friction between the floor and the box? Additional Materials Reading

Answer 1

round your answer as needed

a here a here F- f = ma where F = applied force f = friction force =) Fi- f = ma, F- f = ma, - F - F = m (9,-0₂) =) M= F-F2 a, az We have F = 72N,A, = 0 46 m/s2 f = 79 N 92 = 0.64 172-79 0-46- 0.64 m = 38.8888 8 kg ni 3 9 kg 6) friction foce s n= ) 38.89 f = sing a = Filing =ma, Fi-mai ng 72-38-89 x 0.46 38.89 x 9.81 = 0.14 18 323 is a 0. 142