Question

50. Two boxes A and B, are connected to the ends of a light vertical cord, as shown in figure 40. A constant upward force F= 80.0 N is applied to box A. Starting from rest, box B descends 12.0 m in 4.00 s. Find the mass of box B if the tension in the cord between the boxes is 36.0 N.

Answer 1

3 6 тв a T

Answer 2

There's a force of 36N upwards which is countered by 9.8M
where M be the mass of box B

From kinematic equations
s = ut + 1/2at^2
12 = 0 + 1/2 a * 4^2
12 = 8a
a = 12/8
a = 1.5 m/s^2
The resultant downwards force causes an acceleration of 1.5 ms^2

So the resultant downwards force = 1.5 M
This gives us final equation
9.8M - 36 = 1.5 M
8.3 M = 36
M = 4.34 kg

Answer 3

The system is accelerating, so we apply Newton's second law to each box and can use the constant acceleration kinematics for formulas to find the acceleration. We now that d'x dt and therefore Figure 2 In one dimension (along the y-axis) when the acceleration is constant equation becomes

F-T g+ a (80.0- 36.0) N (9.8-1.5)m/s = 5.30 kg Note that even though the boxes have the same acceleration they experience different forces because they have different masses 90.0 90.0 45.0 90.0 T'

Answer 4

Let m_A be the mass of the box A and let m_B be the mass of the box B.

Total constant upward force applied to box A is, F_A = 80 N

Total constant upward force applied to box B is, F_BU = 36 N

Starting from rest, box B descends 12 m in 4 s.

Initial velocity of the box B is, u_B = 0 m/s

s = 12 m

t = 4 s

From the equation of motion,

s = u_Bt + (1/2)* a_Bt²

12 = 0 * 4 + 0.5 * a_B * 4 * 4

a_B = 12 / (0.5 * 4 * 4)

a_B = 1.5 m/s²

Total downward force on box B is, F_BD = m_B*g

Thus, total accelerating force on B is,

F = F_BD - F_BU

m_B a_B = m_B*g - 36

m_B (g - a_B) = 36

m_B = 36 / (9.8 - 1.5)

m_B = 4.337349398 kg = 4.34 kg