Question

Gravel Box Ms = 0.700 MK = 0.400 Concrete A 50.0-kg bucket of concrete hangs from a cable, and the other end of the cable is attached to a 70.0-kg box, as shown. A 30.0-kg bag of gravel is sitting on top of the box. The coefficients of friction between the box and roof are shown. Initially, nothing is moving. a. Draw separate free body diagrams for (i) the gravel, (ii) the box, and (iii) the bucket. b. Find the friction force on the bag of gravel. c. Find the friction force on the box. Now imagine that the gravel is removed. d. Use Newton's Laws on the moving system to find the acceleration of the bucket. e. Use energy conservation to find the speed of the bucket after it has fallen by 2.0 meters. f. Use your acceleration from part (d) and kinematics to find the speed of the bucket after it has fallen by 2.0 meters. Your answer should agree with part (e).

Answer 1

Nfloor NBox. A -gloor Solnay Gravel (f=0) Box HT Concente Bucket Ifs=0 maig meg on tog Moox. b) [gnitially nothing is moving so, fs=o. c) while frloor friction Box due to floor = T (as Box is not moving ffloor meg as Concrete Bucket is also hot moving so T=meg. i Frloor mcg 50x 9.8=4gON. -490N d BOX -->T (Gravel is removed) Ffloor nog N floor = maig (mig) T-Feloor mba (a- acceleration) But floor applied Kinetic friction this time. T- Uk Nfloor = mba T- 04x50x9.8=70a T- 274.4 =70a T-70a 274.4 Adding ③ Nfloor (Bucket! Ja A Mebg Q=1.8 ms² T=400017N Mobig-T = Moba 400-T = 50a - 50a-T =-40 CD

ey Using Conservation of Energy of the system. Work done by all the force, on Bucket = change in its kinetic ngh - TX2 I my 2 50x9.8 x 2400 17x2 =25v2 Energy 561228 36x9-889- 40847//2542 >= 2.68 ms S=2m. ty a = 1.8 m/s², uzo, v = ? v² =4²+2as V² = 2x1.8x2 v = √72 2.68 ms -