Question

8) A car is pushing a large truck, such that both accelerate. Which is true? A) The contact force the car exerts on the truck is larger than the contact force the truck exerts on the car. B) The contact force the truck exerts on the car is larger than the contact force the car exerts on the truck. C) The non-zero contact force of the truck on the car is equal to the contact force of the car on the truck. D) The contact forces between the car and truck are both zero. E) The contact force of the truck on the car is zero. 9) You are pulling on the rope with a tension of 6 N. The first mass is 2 kg and the sccond is 4 kg, and both arc sliding on a fricitonless surface. What is the tension in the rope connecting the blocks? A) 2 N B) 6 N C)4N D) 3 N E) ON 10) A 7.0-kg object is acted on by two forces. One of the forces is 10.0 N acting toward the east. Which of the following forces is the other force if the acceleration of the object is 1.0 m/s2 toward the east? A) 12 N east B) 7.0 N west C) 3.0 N west D) 9.0 N West B) 6.0 N east 11) A 6.0-kg block is at rest. A net force acts in one dimension. It increases linearly from zero to 8.0 N in 4.0 s, then back to zero in another 2.0 s. What is the final speed of the block? A) 8.0 m/s B) 5.3 m/s ) 3.3 m/s D) 4.0 m/s E) 11 m/s 12) A wooden block A of mass 4.0 kg slides on a frictionless table when pulled using a massless string and pulley array by a hanging box B of mass 5.0 kg, as shown in the figure. What is the acceleration of block A as it slides on the frictionless table? A LOCOS Frictionless A) 2.7 m/s2 B) 4.1 m/s2 C) 3.1 m/s2 D) 3.5 m/s2 E) 2.2 m/s2

Answer 1

solution: FCT IT For: Force on car due to truck. FTC: Farce on truck due to car. By Newtous thind law, Frel=lFctl correct option is c) AFN2 : pfni ForF=6N m,- 2 kg mag mig My = 4kg F-T= m, a, Tema = F = (M+M₂) a → a=- To me a=un. Answer : c) - -Im. 10 ) Enemg F it Fi=10.ON 1 f. f = ma mag R 10-f₂ = 7 F₂ = 3.00 Answer c) 3.on weat.

( с 2:0 - о 1) м – 6. oko. V; =0 m/s fo+ (-) t = 2-ot F, 9-о + 4.o) - 8.0 - 4.0 (t-40) F, 29.0 — Фон Impulse given to the block. Áp = ffat б) - tai • Гечо -ьондан С. (1) 6-0 4-0 - , - 4 ot 9.0 t2 +0 То БР - 16.0 + +8.0 — 2.0 (30 – 16 ) = 24-оку м), m (v-v:) = 24 o o 6. o (v-о) = 2.4 -0 Final velocity Uf = 4.0 m/s Answer D). pFNA MA = 4.0kg му 9-0 kg 12) 2T мА9 Mo9 МОЯ - Т2 MB - Темрар 2 А 2 хв от 2 м-бAP 2.Ад - до 2 м 9 - м, е ов) + MA-0А - ИА ( MA +4 м)

AA = 2MBg MA +4 MB . (2) (5.0) (9.81) 4.0+ 20.0 QA = 4.0875 m/s2 = 4.1 m/s2 Answer B)