King Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 4 Section 10 A 1.50-kg block slides down a frictionless 30.0 incline, starting from rest. What is the work done by the net force on this block after sliding for 4.00 s? The net force acting on the block is the horizontal component of gravity, which is in the same direction as the displacement. The acceleration of the block is a = g x sinθ = 9.80 x sin 30.0 o = 4.90 m/s vf = v o + at = 0 + (4.90 x 4.00) = 19.6 m/s Wnet = K = K f K i = [½ x 1.50 x (19.6) ] 0 = 88 J Another solution: Distance d = v o t + ½ at = 0 + [ ½ x 4.90 x 16.0] = 39. m Wnet = m x g x d x cos φ = 1.50 x 9.80 x 39. x cos 60.0 o = 88 J ***************************************************************************** Section 11 A particle moves in the xy plane from the point (0, 1.0) m to point (4.0, 5.0) m while being acted upon by a constant force F = 4.0 î + 3.0 ĵ + 4.0 kˆ (N). What is the work done on the particle by this force? Displacement: d = r f r = 4.0 î + 4.0 ĵ (m) i Work: W = F d = (4.0 x 4.0) + (3.0 x 4.0) = 16 + 1 = 8 J ***************************************************************************** Section 1 A person lifts a 0.50-kg cup of water a distance of 0.84 m vertically up at constant speed. What is the work done on the cup by the person? K = W net = W g + W a Since the cup is raised at constant speed: K = 0 Thus, W net = W g + W a = 0 W a = W g = (m x g x d x cos φ) = (0.50 x 9.8 x 0.84 x cos 180 o ) = 4.1 J

King Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 5 Section 10 A 5.0-kg box starts up a 30 o incline with 150 J of kinetic energy. How far will it slide up the incline before stopping if the coefficient of kinetic friction between the box and incline is 0.35? Let d be the distance moved on the incline, and h be the vertical height moved by the box W ext = W f = μ k x mg x cosθ x d = (0.35 x 5.0 x 9.8 x cos 30 o ) d = 14.85 d ΔU g = + m x g x h = m x g x d sinθ = 5.0 x 9.8 x d x 0.50 = 4.5 d ΔK = Kf K i = 0 150 = 150 J ΔK + ΔUg = W ext : 150 + 4.5 d = 14.85 d d = 3.8 m ********************************************************************************* Section 11 A 3.00-kg block slides on a rough horizontal table. Just before it hits a horizontal ideal spring its speed is 6.00 m/s. It hits the spring and compresses it 10.0 cm before coming momentarily to rest. If the spring constant is 1500 N/m, how much work is done by friction? ΔU s = + ½ k x = ½ x 1500 x 0.0100 = 7.50 J ΔK = Kf K i = 0 [½ m (v i ) ] = ( ½ x 3.00 x 36.0) = 54.0 J ΔK + ΔU s = W ext : 54.0 + 7.50 = W f W f = 46.5 J ********************************************************************************* Section 1 A projectile is fired from the top of a 50-m high building with a speed of 5 m/s. What will be its speed when it strikes the ground? Ignore air resistance. 1 = Top of the building = Ground Reference = Ground K 1 + U 1 = K + U But U = 0 (on the ground) Thus : K1 + U 1 = K : ½ m (v 1 ) + m g H = ½ m (v ) (v) = (v 1 ) + g H = 65 + ( x 9.8 x 50) = 1605 (m/ v = 40 m/s

King Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 6 Section 10 A ball having a mass of 50.0 g strikes a wall with a velocity of 8.00 m/s perpendicular to the wall and rebounds in the opposite direction with only 50 % of its initial kinetic energy. What is the magnitude of the impulse that acts on the ball while it is in contact with the wall during collision? Solution : K i = ½ m (v i ) = 0.5 x 0.05 x 64 = 1.60 J K f = ½ K i = 0.800 J K f = ½ m (v f ) Thus: v f = 5.66 m/s pi mvi (0.05)( 8.00ˆ) i 0.400ˆ( i N. p f mv f (0.05)( 5.66ˆ) i 0.83ˆ( i N. p p f pi 0.683ˆ( i N. J p 0.683( N. ------------------------------------------------------------------- Section 11 Two 3.0-kg bodies, A and B, collide. Before the collision, the velocity of body A is (1 î + 15 ĵ ) m/s and after the collision body A moves with velocity ( 5.0 î + 10 ĵ ) m/s. Find the magnitude of the impulse delivered to body B. Solution : p Ai mv Ai (3.0)(1ˆi 15ˆ) j 36ˆi 45ˆ( j N. p Af mv Af (3.0)( 5.0ˆi 10ˆ) j 15ˆi 30ˆ( j N. p A p Af p Ai 51ˆi 15ˆ( j N. pb 51ˆi 15ˆ( j N. J p 53( N. B B ------------------------------------------------------------------- Section 1 A.0-kg particle is moving with a velocity of 1 m/s along the positive x direction, while a 3.0 kg particle is moving with a velocity of 5.0 m/s along the positive y direction. Find the magnitude of the velocity of their center of mass. Solution : V com V com 1 ( m1v 1 mv M [(4.8) (3.0) 1 ) [(.0)(1ˆ) i (3.0)(5.0ˆ)] j 4.8ˆi 3.0ˆ( j m / 5.0 1/ ] 5.7m / s

King Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 7 Section10 A rope pulls a.00-kg box on a frictionless surface through a pulley, as shown in the figure below. The pulley has a rotational inertia of 0.0500 kg.m and a radius of 5.0 cm. If the force F is 15.0 N, what is the magnitude of the acceleration of the box? For the box: ma = T 1 (1) For the pulley: I = net I.( a/r) = 1 F = RT 1 RF = R (T 1 F) Thus: T 1 = F (I.a/R ) Back to (1): ma = F (I.a/R ) (m + I/R ) a = F Thus: (.00 + 0.800) a = 15.0 Thus : a = 5.36 m/s ****************************************************************************************** Section 11 The figure below shows a pulley, with a radius (R) of 5.0 cm and rotational inertia (I o ) of 0.0050 kg.m, that is suspended from the ceiling. A rope passes over it with a.0 kg block attached to one end and a 3.0 kg block attached to the other. When the speed of the heavier block is 4.0 m/s, what is the total kinetic energy of the pulley and blocks? Let m 1 =.0 kg and m = 3.0 kg, R = radius of pulley, Io = The angular speed of the pulley: = v/r = 4.0/0.050 = 80 rad/s. Total kinetic energy: K tot = K 1 + K + K pulley K tot = (½ m 1 v ) + (½ m v ) + (½ I o ) K tot = ½ (m 1 + m 1 ) v + (½ I o ) K tot = (0.5 x 5.0 x 16) + (0.5 x 0.0050 x 6400) = 56 J ****************************************************************************************** Section 1 A uniform disk, of radius 0.50 m and mass 15.0 kg, can rotate about its axis. Starting from rest, it reaches an angular velocity of 45.0 rad/s in 10.0 s under the action of a constant torque. (a) What is the magnitude of the angular acceleration of the disk? (b) How many revolutions will it make in this time interval? (c) What is the magnitude of the torque acting on the disk? (d) What is the instantaneous power at the end of this time interval? The rotational inertia of the disk about an axis through its center of mass and perpendicular to its plane is I com = ½ MR. (a) f = i + t. Thus: =( f i )/t = (45.0 0)/10.0 = 4.50 rad/s. (b) = ( i t) + (½ t ) = 0 + (0.5 x 4.50 x 100) = 5 rad = 35.8 rev. (c) I = 0.5 x 15.0 x 0.065 = 0.46875 kg.m. Thus: = I. = 0.46875 x 4.50 =.11 N.m (d) P =. =.11 x 45.0 = 94.9 W

King Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 8 Section 10 A uniform solid sphere of mass 5.0 kg and radius 0.10 m rolls smoothly on a flat horizontal plane. If the speed of the center of mass of the sphere is.0 m/s, what is the total kinetic energy of the sphere? The rotational inertia of the sphere is [I com = MR /5] K = K trans + K rot = [½ M (V com ) ] + [½ I com ] K = [½ M (V com ) ] + [½ (MR /5) x (V com /R) ] K = [½ M (V com ) ] + [⅕ M (V com ) ] K = 0.7 M (V com ) = 0.7 x 5.0 x 4.0 = 14 J **************************************************************************************************** Section 11 At a certain instant of time, a 0.50-kg particle is at the point (.0,.0) m and has a velocity of v = (3.0 î + 4.0 ĵ ) (m/. Find its angular momentum relative to the origin at this moment. (Give your answer in unit vector l ( m ) m( ) (. ).(. ˆ. ˆ ) (. ˆ. ˆ ). ˆ r p r v r v 0 50 0i 0j 3 0i 4 0j 1 0k ( kg.m / s ) **************************************************************************************************** Section 1 A uniform solid disk of mass 5.00 kg and radius 0.500 m rotates about a fixed axis perpendicular to its face and through its center. The angular speed of rotation is 5.00 rad/s. What is the magnitude of the angular momentum of the disk? The rotational inertia of the disk is [I com = ½ MR ] I com = 0.5 x 5.0 x 0.5 = 0.65 kg.m L = I com. = 0.65 x 5.00 = 3.15 (kg.m /

King Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 9 Section 10 A 5.00-m long uniform ladder (with mass m = 15.0 kg) leans against a wall at a point 4.00 m above a horizontal floor as shown in the figure. Assume that the wall is frictionless (but the floor is not). (a) Determine the magnitude of the normal force exerted on the ladder by the wall. (b) Determine the magnitude of the normal force exerted on the ladder by the floor. The forces acting on the ladder are shown. (a) Calculate the torque about O. [W.(L/). cos ] [F nw.l. sin] = 0 F nw = 55.1 N (b) F y = 0 F nf = W = 15 x 9.8 = 147 N ****************************************************************************************** Section # 11 An 80-kg man balances a boy on a massless beam, as shown below. What is the mass of the boy? W m = weight of the man d m = distance of the man from fulcrum W b = weight of the boy d b = distance of the boy from fulcrum Calculate the torque about the fulcrum: W m.d m = W b.d b W b = W m.d m /d b = 80 x 9.8 x 1.0/4.0 = 196 N Mass of the boy = 196/9.8 = 0 kg ****************************************************************************************** Section # 1 A uniform beam, of weight 50 N and length.0 m, carries a 5 N weight at its right end. Another weight W is placed 150 cm from the right end. The system balances horizontally on a support located at a distance of 75 cm from the right end. If the system is in equilibrium, find W. W 1 = 5 N W = weight of the beam = 50 N W =?? F n = normal force at the support Calculate the torque about the support: [W 1.(75)] + [W.(5)] + [W.(75)] = 0 Thus : 75 W = 16875 650 Thus : W = 14 N

King Fahd University of Petroleum and Minerals Physics Department Physics 101 Recitation Term 131 Fall 013 Quiz # 10 Section 10 Planet Mars has a mass of 6.4 10 3 kg and a radius of 3.4 10 6 m. What is the escape speed from this planet? [G = 6.67 10-11 N.m /kg ] GM R 11 3 6. 67 10 6. 4 10 5. 0 10 m / s 3 v esc 6 3. 4 10 ********************************************************************************* Section 11 Planet Mars has a mass of 6.4 10 3 kg and a radius of 3.4 10 6 m. What is the magnitude of the gravitational acceleration (a g ) on the surface of this planet? [G = 6.67 10-11 N.m /kg ] 11 3 GM 6. 67 10 6. 4 10 a g 3. 7 6 R 3. 410 m / s ********************************************************************************* Section 1 Two particles are separated by a distance of 1.0 m. Each particle has a mass of 1.0 kg. What is the magnitude of the gravitational force between them? [G = 6.67 10-11 N.m /kg ] 11 G.m.m. 67 10 1. 0 1. 0 F 1 6. 67 r 1. 0 10 6 11 N