Ch-5 Term 091 Help-Session

CH-5-082 Q15: 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) 9/21/2018

CH-5-072 Q16. 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/s2 Assume acceleration a of mass B downward mA= 70/9.8= 7.14 kg ;mB= 35/9.8= 3.57 kg Then For masses A and B; T=mAa; T-mBg= - mBa Solving for T mAa=mB(g-a) a = mBg/(mA+mB) =(3.57*9.8)/(7.14+3.57) =3.27 m/s2 9/21/2018

CH-5-082 Q17. 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 Fy= N-200 sin20-mg = 0 N= 200 sin20 +mg =200*0.34+25*9.8 = 313 N 9/21/2018

CH-5-081 Q15. Two masses m1 = 10 kg and m2 = 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= 1m/s2 FBD of 10 kg mass T= 10*a= 10*1.0=10 N 9/21/2018

CH-5-081 Q16. 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/s2 Assume acceleration a of 10 kg mass downward Then for 10 kg mass ; T-10*g=-10*a; T=10g-10a But from 5 kg mass T= 5a then 5a=10g-10a or 15 a= 10 g Then a= 10g/15= (10*9.8)/15 a = 6.53 m/s2 9/21/2018

CH-5-081 Q17. A 70 kg man stands in an elevator that is moving downward at constant acceleration of 2.0 m/s2. The force exerted by the man on the elevator floor is (Ans: 546 N down) N-mg=-ma N= mg-ma = m(g-a) = 70 (9.8-2)= 546 N Force exerted by man equal and opposite to N 9/21/2018

CH-5-072 Assume acceleration a of mass m1 upward Then For mass m1; Q13.Two blocks of mass m1= 24.0 kg and m2, 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 m2. (Ignore friction) (Ans: 40.0 kg) Assume acceleration a of mass m1 upward Then For mass m1; T-m1g=m1a; a=(T-m1g)/m1 a =(294-24*9.8)/24= =2.45 m/s2 Then acceleration a of mass m2 is downward Then for mass m2; T-m2g=-m2a; T=m2g-m2a Then m2=T/(g-a)=294/(9.8-2.45) =40.0 kg 9/21/2018

CH-5-072 Q16.: A 5.0 kg block is lowered with a downward acceleration of 2.8m/s2 by means of a rope. The force of the block on the rope is:(Ans:35 N, down) Q14.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) |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 F1=aj; F2=-ai; R=F1+F2 Direction of R= =tan-1(-a/a)=-45° 9/21/2018

CH-5-072 Net force Fnet along x-axis Then Fy=0 i.e F1sin60=F2sin30 Q17.:Two students are dragging a box (m=100 kg) across a horizontal frozen lake. The first student pulls with force F1=50.0 N, while the second pulls with force F2. The box is moving in the x-direction with acceleration a (see Fig. 3). Assuming that friction is negligible, what is F2? (Ans: 86.6 N) Net force Fnet along x-axis Then Fy=0 i.e F1sin60=F2sin30 F2=F1sin60/sin30 =50xsin60/sin30=86.6 N 9/21/2018

CH-5-071 F1= (6i+8j) N |F1|=(36+64)=10 N Q14. Only two forces act upon a 5.0 kg box. One of the forces is F1= (6i+8j) 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) 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/s2. What is the net force on block B? (Ans:12 F1= (6i+8j) N |F1|=(36+64)=10 N Since box is moving at a constant velocity i.e Fnet=0=F1+F2 |F2|= |F1|=10 N Acceleration a=Fnet/mA+mB mB=(Fnet/a)-mA = =(20/2)-4=6 kg FB=2x6=mBa=12 N 9/21/2018

CH-5-071 Q16.: 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: 2000N) Q15. 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/s2? (Ans: 0) T-mg=-ma T=mg-ma=3000-[(3000/9.8)x 9.8]=0 |Fnet|= mass x acceleration a from t = 15 s to t = 20 s a = (0-10)/(20-15)=-2 m/s2 | Fnet|=|ma|=1000x2=2000 N 9/21/2018