Question

A box of mass 40 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is µk = 0.40. A woman pushes against the box with a force of 250 N that makes an angle of 30° with the horizontal until the box attains a speed of 1.0 m/s.
(a) What is the change of kinetic energy of the box?
(b) What is the work done by the friction force on the box?
(c) What is the work done by the woman on the box?
d) Show that the work-kinetic energy
theorem is satisfied.

Answer 1

40kg mass, m 309 Fcose F F Fsine

Resolve the fo e mto is vartinl amd horizontal components Than Fcoss the component of the fore F Acting horizontaly on the block. 2.50N So F Feos A 250x Los30 Given, coefficient of kinahie piction, M o So, the tietan fone F that will oppo se the motion of the block is Mmg So Fn= Mma - 0.40 X 40 x 9.8 N 156.8 N et fom ce acing F =FF Bn he blkock is So, (2.50 -156.3) 2. 59.7 N 59.7 m/s F Hen ce acceeratiom of the block, a= 40 1.49 m/s Change of kineh'c energy of the block: AKE =Final lainetic ene ray nitial kinetice enargy m m amd O ginem 1.0m/s So, xtoxo) AKE. 4o x1 =20 J

D Using the kine maties eqwation, V2AS we will find he distam e ,s, bravelled 1-0 2x 1.49 x S > m 2x1.49 = 0.34 m So, wonk done by ridhan foncas whre is angle F and s which is 1s0because triction tonce and d'splaument s are opposite in diection between WnFS cos 4 WaFs cosiso S0 = FS =- IS6.8 x o.34 J =-53.31 3 mp lying here work done by the Wonk dene by woman harizontal compo nent o 1he force exerte by the woman where o because F and S ane in WFSosA SAme diruction S WFScos 0" = FS 1 73.6 J (The nat wonk dono on the block is N= WntWH -53.31J 73.61S 2 20 J md in port O We gat AKE. = 20 J Hunce W AKE. 20J/ Wonk enery theorem statas that chamge to the work done on the bady K.E, is equal