Two blocks A and B with mA = 1.7 kg and mB = 0.88 kg are connected by a string of negligible mass. They rest on a frictionless horizontal surface. You pull on block A with a horizontal force of 7.9 N. (a) Find the magnitude of the acceleration (in m/s2) of the blocks. m/s2 (b) Determine the tension (in N) in the string connecting the two blocks. N (c) How will the tension in the string be affected if mA is increased? increase decrease remain the same

A) 3.062 m/s2

B) 2.6946 N

C)  

If ma is increased, acceleration will decrease if the applied force remains fixed

Explanation:

Ma =1.7 kg ( Block A )

Mb = 0.88 kg ( Block B )

Force applied = 7.9 N

a) Calculate the magnitude of acceleration of blocks

F = mass * acceleration

acceleration = Force / mass

mass = ( 1.7 + 0.88 ) = 2.58 kg

Force = 7.9 N

Therefore acceleration = 7.9 / 2.58 = 3.062 N/kg =  3.062 m/s2

B) Tension in the string

Tension in string can be calculated using the relationship below considering the first block

F - T = ma * A

A = 3.062 m/s2

ma = 1.7 kg

F = 7.9 N

Hence Tension (T) = F - ma * A

                               = 7.9 - ( 1.7 * 3.062)

                               = 7.9 - 5.2054 = 2.6946 N

C)  If mA is increased, acceleration will decrease if the applied force remains fixed

6/12=1/2

step-by-step explanation:

f=ma newton

f=100x2.5=250n