Cypress Creek High School IB Physics SL/AP Physics B MP2 Test 1 Newton s Laws. Name: SOLUTIONS Date: Period:

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1 Nme: SOLUTIONS Dte: Period: Directions: Solve ny 5 problems. You my ttempt dditionl problems for extr credit. 1. Two blocks re sliding to the right cross horizontl surfce, s the drwing shows. In Cse A the mss of ech block is 3.0 kg. In Cse B the mss of block 1 (the block behind) is 6.0 kg, nd the mss of block 2 is 3.0 kg. No frictionl force cts on block 1 in either Cse A or Cse B. However, kinetic frictionl force of 5.8 N does ct on block 2 in both cses nd opposes the motion. For both Cse A nd Cse B determine () the mgnitude of the forces with which the blocks push ginst ech other nd (b) the mgnitude of the ccelertion of the blocks. Cse A (Block A) Cse B (Block B) - x + x - x + x FNA FNB Block A FBA= -FAB FAB f k Block B m Ag m Bg Combining the bove results, we find Then b 3.0 kg5.8 N 3.0 kg 3.0 kg2.9 N kg5.8 N 6.0 kg 3.0 kg3.9 N Note: The positive vlue for the ccelertion mens tht the direction chosen for in the free-body digrm is correct. A negtive vlue would men tht is in the opposite direction to the one chosen

2 2. In the drwing, the rope nd the pulleys re mssless, nd there is no friction. Find () the tension in the rope nd (b) the ccelertion of the 10.0-kg block. (Hint: The lrger mss moves twice s fr s the smller mss.) SOLOTION: + y FN 1 - x + x T T m 1g = 10.0g m 1 = 10.0 kg T - y x = 2y v 1 = x y = 2 t t = 2v 2 1 = v 1 v 2 = 2 = 2 2 t t 1 = 22 m 2g = 3.00g m 2 = 3.00 kg b Combining the two equtions for T, we find m s1.37 m s kg1.37 m s 13.7 N

3 3. A smll sphere is hung by string from the ceiling of vn. When the vn is sttionry, the sphere hngs verticlly. However, when the vn ccelertes, the sphere swings bckwrd so tht the string mkes n ngle of with respect to the verticl. () Derive n expression for the mgnitude,, of the ccelertion of the vn in terms of the ngle nd the mgnitude g of the ccelertion due to grvity. (b) Find the ccelertion of the vn when 10.0 o. (c) Wht is the ngle when the vn moves with constnt velocity? y T cosθ T T sinθ θ x -direction: 0 cos -direction: sin () Dividing the -direction eqution by the -direction eqution, sin cos tn (b) If 10 tn m/s (c) If the velocity of the vn is constnt, then 0 tn tn 0

4 4. The drwing shows three objects. They re connected by strings tht pss over mssless nd friction-free pulleys. The objects move, nd the coefficient of kinetic friction between the middle object nd the surfce of the tble is () Wht is the ccelertion of the three objects? (b) Find the tension in ech of the two strings. T1 FN T2 T1 T2 f k m1g = 10.0g m2g = 80.0g m 1 = 10.0 kg m 2 = 80.0 kg m 3 = 25.0 kg Combining the equtions for nd bove, we find m s m3g = 25.0g b N N

5 5. A block of mss 2.0 kg rests on top of nother block of mss 10.0 kg tht itself rests on top of frictionless tble (see figure below). The lrgest frictionl force tht cn develop between the two blocks is 16 N. () Drw seprte free-body digrms showing ll the forces cting on ech block. (b) Clculte the ccelertion of the blocks. (c) Clculte the lrgest force, F, with which the bottom block cn be pulled so tht both blocks move together without sliding on ech other. () F N1 f s F N2 f s m 1g = 2.0g m 2g = 10.0g (b) Applying Newton s 2 nd lw to the 2.0-kg block yields N N kg (c) Applying Newton s 2 nd lw to the 10.0-kg block yields 10.0 kg8.0 N 16 N 96 N kg

6 6. A mn of mss m stnds in n elevtor. Find the rection force from the elevtor floor on the mn when () the elevtor is stnding still; (b) the elevtor moves up with constnt speed ; (c) the elevtor ccelertes down with constnt ccelertion ; (d) the elevtor ccelertes down with ccelertion. (e) Wht hppens when the elevtor moves down with? F N W = mg () 0 (b) 0 (c) (d) 0 (e) 0. The mn rises nd hits the top of the elevtor.

7 7. An 81-kg bsebll plyer slides into second bse. The coefficient of kinetic friction between the plyer nd the ground is () Wht is the mgnitude of the frictionl force? (b) If the plyer comes to rest fter 1.6 s, wht ws his initil velocity? () (b) kg9.80 m/s 389 N 389 N 4.8 m/s 81 kg m s 1.6 s 7.7 m/s

8 8. The drwing shows lrge cube (mss = 25 kg) being ccelerted cross horizontl frictionless surfce by horizontl force P. A smll cube (mss = 4.0 kg) is in contct with the front surfce of the lrge cube nd will slide downwrd unless is P sufficiently lrge. The coefficient of sttic friction between the cubes is Wht is the smllest mgnitude tht P cn hve in order to keep the smll cube from sliding downwrd? FN1 f s f s - FC P FC mg = 4.00g Mg = 25.0g Lrge Cube 25.0 For the system consisting of the two blocks, Substituting nd into the eqution of motion for the lrge cube, we find: N 29.0 kg Smll Cube N N 13.8 N kg Alterntively, we cn find from the eqution of motion of the smll cube before the substitution N 13.8 N kg 4.00 kg

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