Ch5 Term 091 HelpSession CH5 082 Q 15

Ch-5 Term 091 Help-Session

CH-5 -082 Q 15: You stand on a spring scale on the floor of an elevator. The scale shows the highest reading when the elevator: (Ans: moves downward with decreasing speed) 9/16/2020 2

CH-5 -072 Q 16. A 70 N block A and a 35 N block B are connected by a string, as shown in Fig 3. If the pulley is massless and the surface is frictionless, the magnitude of the acceleration of the 35 N block is: A) 3. 3 m/s 2 Assume acceleration a of mass B downward m. A= 70/9. 8= 7. 14 kg ; m. B= 35/9. 8= 3. 57 kg Then For masses A and B; T=m. Aa; T-m. Bg= - m. Ba Solving for T m. Aa=m. B(g-a) a = m. Bg/(m. A+m. B) =(3. 57*9. 8)/(7. 14+3. 57) =3. 27 m/s 2 9/16/2020 3

CH-5 -082 Q 17. When a 25. 0 kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20◦ below the horizontal, the magnitude of the normal force of the floor on the crate is: A) 313 N 9/16/2020 Fy= N-200 sin 20 -mg = 0 N= 200 sin 20 +mg =200*0. 34+25*9. 8 = 313 N 4

CH-5 -081 Q 15. Two masses m 1 = 10 kg and m 2 = 20 kg are connected by a light string and pulled across a frictionless surface by a horizontal force F = 30 N as shown in Figure 2. Find the tension in the string. (Ans: 10 N) Fnet=( m)a 30=(10+20)a a= 1 m/s 2 FBD of 10 kg mass T= 10*a= 10*1. 0=10 N 9/16/2020 5

CH-5 -081 Q 16. A 5. 0 -kg block and a 10 -kg block are connected by a light string as shown in Figure 3. If the pulley is massless and the surface is frictionless, the magnitude of the acceleration of the 5. 0 kg block is (Ans: 6. 5 m/s 2 Assume acceleration a of 10 kg mass downward Then for 10 kg mass ; T-10*g=-10*a; T=10 g-10 a But from 5 kg mass T= 5 a then 5 a=10 g-10 a or 15 a= 10 g Then a= 10 g/15= (10*9. 8)/15 a = 6. 53 m/s 2 9/16/2020 6

CH-5 -081 Q 17. A 70 kg man stands in an elevator that is moving downward at constant acceleration of 2. 0 m/s 2. The force exerted by the man on the elevator floor is (Ans: 546 N down) 9/16/2020 N-mg=-ma N= mg-ma = m(g-a) = 70 (9. 8 -2)= 546 N Force exerted by man equal and opposite to N 7

CH-5 -072 Q 13. Two blocks of mass m 1= 24. 0 kg and m 2, respectively, are connected by a light string that passes over a mass less pulley as shown in Fig. 2. If the tension in the string is T = 294 N. Find the value of m 2. (Ignore friction) (Ans: 40. 0 kg) Assume acceleration a of mass m 1 upward Then For mass m 1; T-m 1 g=m 1 a; a=(T-m 1 g)/m 1 a =(294 -24*9. 8)/24= =2. 45 m/s 2 Then acceleration a of mass m 2 is downward Then for mass m 2; T-m 2 g=-m 2 a; T=m 2 g-m 2 a Then m 2=T/(g-a)=294/(9. 8 -2. 45) =40. 0 kg 9/16/2020 8

CH-5 -072 Q 14. Two horizontal forces of equal magnitudes are acting on a box sliding on a smooth horizontal table. The direction of one force is the north direction; the other is in the west direction. What is the direction of the acceleration of the box? (Ans: 45° west of north) F 1=aj; F 2=-ai; R=F 1+F 2 Direction of R= =tan-1(-a/a)=-45° 9/16/2020 Q 16. : A 5. 0 kg block is lowered with a downward acceleration of 2. 8 m/s 2 by means of a rope. The force of the block on the rope is: (Ans: 35 N, down) |force of block on the rope | = |force of rope on the block |=T Calculate the tension T T-mg=-ma T=m(g-a)=5(9. 8 -2. 8) =35 N 9

CH-5 -072 Q 17. : Two students are dragging a box (m=100 kg) across a horizontal frozen lake. The first student pulls with force F 1=50. 0 N, while the second pulls with force F 2. The box is moving in the x-direction with acceleration a (see Fig. 3). Assuming that friction is negligible, what is F 2? Net force Fnet along x-axis Then Fy=0 i. e F 1 sin 60=F 2 sin 30 F 2=F 1 sin 60/sin 30 =50 xsin 60/sin 30=86. 6 N (Ans: 86. 6 N) 9/16/2020 10

CH-5 -071 Q#13: A constant force F of magnitude 20 N is applied to block A of mass m = 4. 0 kg, which pushes block B as shown in Fig. 5. The block slides over a frictionless flat surface with an acceleration of 2. 0 m/s 2. What is the net force on block B? (Ans: 12 Q 14. Only two forces act upon a 5. 0 kg box. One of the forces is F 1= (6 i+8 j) N. If the box moves at a constant velocity of v= (1. 6 i +1. 2 j )m/s, what is the magnitude of the second force? ( Ans: 10. N) F 1= (6 i+8 j) N |F 1|= (36+64)=10 N Since box is moving at a constant velocity i. e Fnet=0=F 1+F 2 |F 2|= |F 1|=10 N Acceleration a=Fnet/m. A+m. B=(Fnet/a)-m. A = =(20/2)-4=6 kg FB=2 x 6=m. Ba=12 N 9/16/2020 11

CH-5 -071 • Q 15. An elevator of mass 480 kg is designed to carry a maximum load of 3000 N. What is the tension in the elevator cable at maximum load when the elevator moves down accelerating at 9. 8 m/s 2? (Ans: 0) T-mg=-ma T=mg-ma=3000 -[(3000/9. 8)x 9. 8]=0 9/16/2020 • Q 16. : A car of mass 1000 kg is initially at rest. It moves along a straight road for 20 s and then comes to rest again. The velocity – time graph for the movement is given in Fig. 6. The magnitude of the net force that acts on the car while it is slowing down to stop from t = 15 s to t = 20 s is: (Ans: 2000 N) |Fnet|= mass x acceleration a from t = 15 s to t = 20 s a = (0 -10)/(20 -15)=-2 m/s 2 | Fnet|=|ma|=1000 x 2=2000 N 12