Q1: The mass m1 = 1.0 kg is resting on an inclined surface which is makin 35° with the horizontal. Assuming that m2 also has a mass of 1.0 kg and that the whole assembly what is the static friction force? equilibrium A) ON B) 4.18 N C) 5.62 N D) 8.03 N m1 =- fs-mgsinQ+ T=0 E) other Fret = 0 Thet = 0 TIR - T2R = 7x= 0 M2: T-mg=0 +5=mg-mg sino = mg (1-sing)=4.18N fr 035° - 1 kg V U
Q2: Consider the system in the figure: two masses are connected by a very light (massless) string that passes over a pulley. The pulley has radius R, mass M and moment of inertia I = 1/2 M R2. The string does not slip and the pulley rotates without friction. The masses have the following relation: m1= 3m, m2 = m, M = 2m. 2= T1 (a) How do the two tension forces compare? 1 TIR- TER= IX RT=T& R+TX-x= T1-T2=ma V T2 A) T1 < T2 B) T1 = T2 @Ti> 12 m 1 ?=/MR2=mR2 m2 (b) What is the linear acceleration of m2, expressed in terms of the acceleration due to gravity? m,q -T1=m, a A) 0.5 g B) 0.25 g C) 0.33 g D 0.4 g E) 0.67 g 3mg -3ma =T, m29-T2 = - m2Q mg-[2 =- mx->T= m(g+a) 3mg-3ma-ma-ma=ma 2 mg = 5 ma a=?g
Q3: A small 100-g mass is attached to a massless string and moves along a circular path with an initial radius Ri. The other end of the string passes through a small hole at the centre of the table. By pulling the string through the hole we are able to shorten the radius of the mass's circular orbit. Assume that there is no friction between the mass and the table. We now shorten the string to Rf = 2/3 Ri. By what factor f did the kinetic energy change, i.e. if we write K?=f.K; what is f? K == Iw2 1 I = mg2 V R m angular momentum A) 0.44 B) 0.67 C) 1.5 D)2.25 E) other 2 (3) 2 11 9 mur "Li =Lf Ii W1 = If W+ Wp = Wi 4 2 Up =/If WE kg= /5g (?)'W12 =?? Ii us? MR2 12 Ki=(S)hì If =ki = m(2)2 R2 2. 25kg