Question

Tallula is sitting on a toboggan. Bob

pushes on Tallula with a force of 400 N due

South. The force of friction on the toboggan

is 260 N due North. The mass of Tallula and

her toboggan is 65.0 kg.

(a) Determine her acceleration. Give the mag-

nitude and direction.

(b) When Tallula has achieved a speed of

3.00

m

s

, Bob stops pushing. How long af-

ter Bob stops pushing will Tallula come to

a stop?

(c) What distance will Tallula travel after

Bob stops pushing

Answer 1

Ж+V сум Net ОЮ is E F S Tattuttt Техдо tallula TEE т 10 Фx eakck & a quictional force bob - 4х л Р. 260— Что Труд- г. - Чом р ma а - Е (7 - 46 - 1 4б 6S - 2. 13 ~45“ CS Scanned with CamScanner

{ |oj = 2, 15 ups? Dir ection is towards towards South “9_oxy 1. е. b) as ло speed of talluta Bob stor pushing is us 3ups Bob. stops pushing Tallila і | beae Ои in south direction but the frictional force и ишло stop Tallula grictional forul лиошоу force will 10 = 260 26o N ма a> 266 4 ups? 6s Thir а(( сосан би will be in opposite direction ю о мону Talula сигарх. final 1) Доа f tallula is u=o 5. نو at + — — се a - о - 2 a since the ассе од «Нои шо Hои oppo se the C - -Чщds 2 t- 3 2 0.25 S Ч CS Scanned with CamScanner

c) e2-42=zas o-32 -284 sa s= ola S = 1. 125 m CS Scanned with CamScanner