A)a block of mass m=6.40 kg is suspended as shown in the diagram below.assume th
ID: 1329265 • Letter: A
Question
A)a block of mass m=6.40 kg is suspended as shown in the diagram below.assume the pulley to be frictionless and the mass of the strings to be negligible.If the system is in equilibrium, what will be the reading of the spring scale in newtons? B) Two blocks each of mass m= 6.40 kg are connected as shown in the diagram below. Assume the pulley to be frictionless band the mass of the strings to be negligible.If the system is in equilibrium, what will be the reading of the spring scale in newtons?C) A block of mass m=6.40 kg is in equilibrium on an incline plane of angle theta=26.0 degrees when connected as shown in the diagram below. assume the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons? wtoo's Laws Page 4 of I (a) A block of massm 6.40 kg is suspended as shown in the diagram below Assume the pulley t frictionless and the mass of the strings to be negligible. 15 the system is in equilibrium reading of the spring scale in newtons? to be what will be the (b) Two biocks each of mass m 6,40 kg are connected as shown in the diagram below. Assume the pulley to be frictionless and the mass of the strings to be neghigible. If the system is in equlibrium, what will be the reading of the spring scale in newtons? (c) A block of mass m·6,40 kg is equabrium on an inchne plane of angle -26.0° when connected as shown in the diagram below. Assume the mass of the strings to be system is in equilibrium, what will be the reading of the spring scale in nentons? negligible. If the Aoditional Beadna
Explanation / Answer
a)
spring will read the tension T
= m*g
=6.4*9.8
= 62.72 N
b)
spring will read the tension T
= m*g
=6.4*9.8
= 62.72 N
c)
spring will read the tension T
= m*g*sin(thetha)
=6.4*9.8 * sin(26)
= 27.5 N