Part A - In the Top configuration, what is the block's initial Kinetic Energy? Vi Lo Part B - In the bottom configuration, what is the spring's potential energy? m Part C - In the bottom configuration, what is the block's speed vf? Vf L Provide Feedback m In the Initial configuration (top drawing), on a frictionless horizontal surface, a 2.2 kg block has an initial speed vi = 4.9 m/s and encounters a spring. The spring constant is k = 398.0 N/m . Its natural length is Lo = 0.29 m . In the bottom drawing, the block compresses the spring to a new length L = 0.10 m .Part A - Find Vunder, the solid object's volume that is UNDER the water surface. A solid object floating in water. Part of the object's volume is under the surface (denoted as Vunder). If you enter a regular number, keep 6 digits after the decimal point, in m3 If you use scientific notation such as 1.234 .10-3 , keep 3 digits after the decimal point. VO ? Vunder = m3 The mass of the solid object is m = 3.9530 kg and its total volume is V = 6.70x10-3 m . The density of water is Pw = 1000.0 kg/m , and the Submit Request Answer magnitude of the gravitational acceleration is g = 9.80 m/$2Review Part A - Find the angular acceleration with a proper sign. Keep 3 digits after the decimal point, in rad/s2 Rotational axis IVO AEQ angular acceleration a = rad/ $2 A round solid disk rotates counterclockwisely about Submit Request Answer its center. Its initial angular velocity is wi = 16.4 rad/s . The disk slows down to a final angular velocity of wf = 8.0 rad/s in t = 3.2s . Part B - What is the angular displacement in the slowing down process? The disk has a mass m = 2.4 kg and a radius of R = 0.49 m . Its moment of inertia is - mR2. Keep 2 digits after the decimal point, in rad/s2 IVO AED ? angular displacement 40 = radPart A - Relative to this rotational axis (RIGHT support), which of the following contains the correct torques of the two forces, N1 and NV2including proper signs? N, and N2 represent the magnitudes of these two forces N2 Lever arm calculations are in the hints. weight If you make ONLY ONE attempt in a hint, even if it is wrong, you don't lose partial credit. Weight block plank In a hint, if you make 2 attempts and both are wrong, or if your "request answer", you lose partial credit. View Available Hint(s) As shown in the figure, a uniform plank rests on two O torque of N1 = 0 , torque of N2 = +L *N2 supports in equilibrium. The plank's center of mass is at the midpoint of the plank. A block is placed on the left end of the plank. The magnitude of the gravitational O torque of N1 = 0, torque of N2 = - L *N2 acceleration is g = 9.80 m/sz . torque of N1 =+D*N1 , torque of N2 = 0 L = 6.00 m , D = 4.80 m , mplank = 35.5 kg , mblock = 6.0 kg O torque of N1 = +(L- D)*N1 , torque of N2 = +L *N2 Choose the RIGHT support as rotational axis. It is good practice to mark the rotational axis on the drawing. O torque of N1 = - (L- D)*N1 , torque of N2 = - L *N2 In the following steps, you will find the magnitude of the two normal forces, N, and N2 torque of N1 = - D*N1 , torque of N2 = 0 O torque of N1 = +(L- D)*N1 , torque of N2 = - L *N2 O torque of N1 = - (L- D)*N1 , torque of N2 = + L *N2 Submit Part B - Relative to this rotational axis (RIGHT support), what is the torque of the weight of the PLANK? Include a proper sign. Lever arm calculation is in the hints. If you make ONLY ONE attempt in a hint, even if it is wrong, you don't lose partial credit. In a hint, if you make 2 attempts and both are wrong, or if your "request answer", you lose partial credit. Torque: Keep 2 digits after the decimal point, in Nmcapped IF, Part A - In the LEFT arm, what is the pressure at the surface of water right below the Left cap, Pleft cap ? F1 capped Area Area oil h=? Keep 1 digit after the decimal point, in Pascals. A water IVO AEQ 6 03 ? As shown in the figure, water is first poured into a U-shaped tube, then some unknown fluid is added Right below the LEFT cap Pleft cap Pa = to the RIGHT arm. There is no intermixing between the two liquids. Then both ends are fitted with tight caps of the same cross-sectional area = 0.014 m Submit Request Answer Force F1 = 9.9 N is exerted on the Left cap, and F2 = 7.6 N is exerted on the right cap Part B - in the RIGHT arm, what is the pressure at the oil surface right below the Right cap, Prightcap ? When static equillibrium is established, in the Left arm h1 = 0.11 m . The density of water is p = 1000.0 kg/ms Keep 1 digit after the decimal point, in Pascals. The density of oil is poil = 910.0 kg/m . The magnitude of the gravitational acceleration is g = IVO AEd ? 9.80 m/s2 below the RIGHT cap Prightcap Pa = Submit Request Answer Part C - In the LEFT arm, what is the pressure at point A, (in water along the horizontal dashed line), PA ? Keep 1 digit after the decimal point, in Pascals. AE In LEFT arm, point A PA Pa = Submit Reallest AnswerCheck that the angle in your calculator is in "degrees", sin30 = 0.5 later initial Part A - The magnitude of the normal force FN has been calculated ro be FN = 9.70 N . Calculate the work done by the normal force. The displacement is d = 11.0 m to the LEFT. Mass m Mass m Keep 2 digits after the decimal point. Moves to the LEFT IVO AEQ The force F2 pulls the suitcase to the LEFT while the force F1 Work by Normal force = tries to slow it down. The magnitude of F2 = 75.0 N . In the drawing the angle 0 = 61. . . The magnitude of F1 is Fi = 30.00 N . The displacement is d = 11.0 m to the LEFT. Submit Request Answer The suitcase' mass is m = 4.7 kg , The magnitude of the gravitational acceleration is 9.80 m/s . The magnitude of the normal force FN has been calculated ro be FN = 9.70 N . The Part B - Calculate the work done by gravity The displacement is d = 11.0 m to the LEFT. magnirude of the kinetic friction force is fk = 4.85 N . The weight vector, the normal force vector and the kinetic Keep 2 digits after the decimal point. frictional force vector are not explicitly drawn in the figure, because this is an exam. AEd ? Work by gravity = Submit Request Answer Part C - Calculate the work done by force F1 The displacement is d = 11.0 m to the LEFT. Keep 2 digits after the decimal point. ?Part D - Calculate the work done by force F 2 The displacement is d = 11.0 m to the LEFT. later initial Keep 2 digits after the decimal point. F IVO ? Mass m Mass m Work by F2 = Moves to the LEFT Submit Request Answer The force F2 pulls the suitcase to the LEFT while the force F1 tries to slow it down. The magnitude of F2 = 75.0 N . In the Part E - The magnitude of the kinetic friction is fk = 4.85 N . Calculate the work done by force f k drawing the angle 0 = 61. " . The magnitude of F1 is F1 = The displacement is d = 11.0 m to the LEFT 30.00 N . The displacement is d = 11.0 m to the LEFT. The suitcase' mass is m = 4.7 kg , The magnitude of the gravitational acceleration is 9.80 m/s . The magnitude of the Keep 2 digits after the decimal point. normal force FN has been calculated ro be FN = 9.70 N . The magnirude of the kinetic friction force is fk = 4.85 N . IVO AEd ? The weight vector, the normal force vector and the kinetic frictional force vector are not explicitly drawn in the figure, because this is an exam. Work by f k = Submit Request Answer Part F - Calculate the Total work done by all the forces. Keep 2 digits after the decimal point. AEd ? Total Work = Submit Request AnswerPart G - The suitcase starts with an initital spped of vj = 3.20 m/s . What is its final speed after being pulled to the LEFT through a displacement of 11.0 m ? later initial Keep 3 digits after the decimal point. F e IVO AEQ ? Mass m Mass m Uf = m/s Moves to the LEFT Submit Request Answer The force F2 pulls the suitcase to the LEFT while the force F1 tries to slow it down. The magnitude of F2 = 75.0 N . In the Part H - New situation: The magnitude of F1 is changed but other forces remain the same. The drawing the angle 0 = 61. . . The magnitude of F1 is Fi = 30.00 N . The displacement is d = 11.0 m to the LEFT. displacement remains the same of d = 11.0 m to the LEFT. If the suitcase is to maintain a constant speed of 3.20 m/s , what is the new work done by the new force F1? The suitcase' mass is m = 4.7 kg , The magnitude of the gravitational acceleration is 9.80 m/s . The magnitude of the normal force FN has been calculated ro be FN = 9.70 N . The Keep 2 digits after the decimal point. magnirude of the kinetic friction force is fk = 4.85 N . The weight vector, the normal force vector and the kinetic frictional force vector are not explicitly drawn in the figure, because this is an exam. AEQ New situation, constant New Work by new speed, F1 = Submit Request Answer Part I - New situation 2: The magnitude of force F1 is changed but other forces remain the same. The displacement remains the same of d = 11.0 m to the LEFT. If the suitcase from the initial speed of 3.20 m/s comes to a stop after the displace of d = 11.0 m to the LEFT, what is the new work done by the new force F1? Keep 2 digits after the decimal point.Part A - What is the power output of the machine that controls the string? Keep 1 digit after the decimal point. Sy: VO AED ? Power = Watt Submit Request Answer An elevator car is being lifted upward by a string at a constant speed of 1.22 m/s . The magnitude of Part B - How much time does the machine (that controls the string) do 500000 Joules of work? the tension force in the string is 4116.00 N . Keep 2 digits after the decimal point.. Unit is Seconds. AED ? time = SOLX = L Ly=0 O L x = 0 Ly = - L Lifting force L O L x = 0 Ly= L onslq to itsq O L x = -L Ly=0 O Lx = Lcos(0) Ly= Lsin(0) O Lx = - L sin(0) Ly= L cos(0) Weight O Lx = - Lcos(0) Ly= L sin(@) A model airplane is making a right turn. Its mass is Submit Request Answer m = 98.0 kg , The magnitude of the gravitational acceleration is 9.80 m/s2 . The speed is 24.56 m/s. The angle 0 = 23.0 0 Part B - Find the magnitude L of the lift Force. Keep 1 digit after the decimal point. Unit is Newtons