My professor gave back my physics test, I am confused on what I missed on the test. Can someone explain to me all of the questions I missed?

1. A 250 g block is dropped onto a vertical spring with a spring constant of k = 2.5 N/cm. The block becomes attached to the spring, and the spring compresses 12.0 cm before momentarily stopping. a) How high was the block dropped from? M= 0.25 kg 0, 25. 9. 81 . 1= 0. 5250. 012 2 k= 2 500 N/m X= 0,12m h = 18 2.4525 ( h 2 0. 734 / b) What is the speed of the block just before it hits the spring? man = 0.25 ( 9. 81) ( 0.12 ) = 0. 294 J 0. 5. 0. 25. V = -(0.294-18)=1 11. 9ke - O.S. K . x = - 0.5. 2500. 0, 12 2 = -185 c) If the speed at impact is doubled, what is the maximum compression of the spring? 0.5 . 0.25. ( 2*11.9 ) 10.5.2500. x -1 0.25 (9.81)x X = 0.238m 22. A 2.00 kg block is pushed against a spring with negligible mass and a force constant of 4.7 N/cm. The spring is compressed to 22.8 cm with the block and the block is then released. The block moves on a plane which has no friction and then on an incline with a slope of 28 degrees with kinetic friction of 0.3. a) What is the speed of the block as it slides along the horizontal surface after having left the spring? F= 4. 7N/Cm zmv? = 1 kx ? - mu TU - Ot - kx 2 2.8 4 , 7 = 34 . $15173 cm/s b) How far along the incline did the block travel before coming to a stop? O +maksin () = 3kx = Kx 2 4. 7 ( 22. 8 ) 2 132, 62618 Cm Zmysing = 2 (2. 9.81. six28) 5 c) While returning back the kinetic friction is different such that the block comes to rest at the bottom of the incline. What is the coefficient of kinetic friction on the downward trip of the block? (Bonus 5 points) d= o. ZZ Fnormal = mig. Cos28 sing 9 6-23.26N : 2. 9.81. 10528 MK = 23,26 20 9.81. (0528 ) - : 0.378 2803. A ball of mass 1.5 kg is fastened to a string 40.0 cm long and fixed at one end and is released when the cord is horizontal. At the bottom of its path, the ball strikes a 3.50-kg steel block initially at rest on a frictionless surface. The collision lasts for 7.25 milliseconds. The impact moves the block which then compresses the spring with spring constant of 2.00 N/cm to 5 cm before coming to rest. a) Find the speed of the block just after collision. V= 1. 5( 9 . 81) ( 0.4 ) O 1 5, 88 J Spring = 2 (200 N/mm) (0.05)=0,5 K = 12mv2 S. 88= 2 (1.S ) V + 0.5 V 2 = 5.38 0.75 -) 12.68 , 1/ b) Find the speed of the ball just before collision? 5.88= 2(1.5) 12- V = /176 1.5 17.8m/s c) Find the speed of the ball just after collision? 3. 5. 2, 68 - (1.5). 2.68= 3. 5. X ip= mv 2.2.64 = 3. 5.V O. Vblock inital + 3. Sublockinitial V= 1.536m's 3 . SVblock + 1.5 Vball c) What is the average force exerted on the ball by the block? Ap= mball ( Val - Vball inital ) 8. 041 Ap = 1.5. (2160 - (2.68)1 0.00725 Ap = 1. 5.5.36. Ap = 8.04 kop mis. F= 1110. 34 N4. A force Fx, shown as a function of distance in the figure, acts on a 5.00 kg mass. The particle starts at x = 0 m with a velocity of 3.58 m/s. a) Determine the total energy in the object when it is at x = 10.0 m. Total energy is the integral Force F in N ( 4 ) ( 8 ) = 16 2 (S)(10) = 25 23-18=5 Position (x in m) 16+(-16)+48+(-16)+25-1 57 J b) At what value of'x' does the particle have maximum energy? What is the value of maximum energy? Force at the highest is at / x= 23 16+ 25= 41 41J c) Determine the speed of the particle at x = 8. F F=ma as- m V + 2 . a . AX - V= V Za . AX 12. 528 36 / /5 V= 12.(9.81) ( 8 ) d) At what value of 'x' does the mass have zero velocity? This is when the the graph crosses the x axis or / x = 4 and 180) 5