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4 years ago

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•• CP Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block AA has mass 15.0

kg, and block BB has mass mm. A constant horizontal force F=60.0 NF=60.0 N is applied to block AA (Fig. P4.38). In the first 5.00 s after the force is applied, block AA moves 18.0 m to the right. (a) While the blocks are moving, what is the tension TT in the rope that connects the two blocks? (b) What is the mass of block B?

1 answer:

Firdavs [7]4 years ago

7 0

Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+ (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

m = 26.67 kg

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