Question

Parts A-E

(25%) Problem 4: A shopper pushes a grocery cart for a distance 22.5 m at constant speed on level ground, against a 47.5 N frictional force. He pushes in a direction 25° below the horizontal. * 20% Part (a) What is the work done on the cart by friction, in joules? Wp= 47.5 X Attempts Remain 20% Part (b) What is the work done on the cart by the gravitational force, in joules? Wg=0 Correct! ► A 20% Part (c) What is the work done on the cart by the shopper, in joules? 20% Part (d) Find the magnitude of the force, in newtons, that the shopper exerts on the cart. 20% Part (e) What is the total work done on the cart, in joules? Wnet * ( 7 8 8 9 9 HOME sino COSO tan() cotan) asino acos atan) acotan() Sinh) cosh() tanho cotanho Degrees O Radians VOL

Answer 1

d=22.5m. Wang for constant speed, accele latino (a)=0 therefore the net furre EFF = ma=0 E 325-Fs ro Where F = Applied force by shopper F. = frictional force = 47.5N Now, I WB 25 - ul.SN=0 F =52000W @ work done by friction (Wp) = -Fs d We -(nr.5N) (22.5m) Wq = -1068.75 TL 6 Work done by gravity (Wg) = 0 (Because no vertical displacement) ③ Work done by shopes (Wshop) = (F 0325) d = Fs d = (17.5N) (22.5m) Wshop = 1068.755 'F=520MIN @ Total work done Wher = We tWg + Wshop = -1068.75 to +1068.755

Answer 2

Horizontal 20:25° of 22.5m Giveni shopper pushe the the grocery cant grocery cars for a distance - 22.5 m frictional force = 18.5 N direction of pushing (o)= 25% below the Horizontal (a) Work done on the cant by fruction The work done work done by by tricte friction is in the opposite direction of motion. So, 0 = 1800 Work done W- fod focos .. Cuts)(225) OSLO = -1068.75 J force = 47.5 N ty gravitational fonce (6) Work done by gravitational fonce : Work done by gravity perpendicula to direction of motivo. so, o = 90° . Wy- Éd cos o = 47.5) (22.5) cos 90° Oj [ cos 90º=0] @ work done on eart by shoppers here, and the cart moves at a constant speed 80, the velocity remains constant Now we renow, lanetic energy a= bare scrice velocity remains same .. linetic energy is also sane before and after pushing. so change in ninetic energy = 0 accordery to work- enengy theorem Net work done a change in tinetic enengy . Whet = 0

cart Now, on the cart, Wnet - Weriction & Wonopper W snopper - Wnet - Wirection - 0 - (-1068.75 ) = 1068.75 T in the work done on the cast by shoppen is 10.68-75J Eshopper Now, Work done = Fod. - Wonopper? Fshopper d cose . Fshoppen Wshopper d coso = 1068.75 3 (22.5) cos 250 252.41 N e) According to work-enengy theonem. Net work done a change in ninetic energy = 0 ) [ Becaure, the cast has a constant speed aso, the kinetic energy is also constant I