Question #118991
In the system shown, a horizontal force Fx = 40 N acts on an object of mass m2 = 8.00 kg. A friction force fk = 5 N acts between object m2 and the table top, and m1 = 2.00 kg. i) Calculate the acceleration of the masses. ii) Calculate the tension in the string.
1
Expert's answer
2020-06-01T14:18:43-0400

Given,

The masses m1=2Kg,m2=8Kgm_1=2Kg,m_2=8Kg and frictional force acting between table and those masses is f=fk=5Nf=f_k=5N .

Horizontal external force is Fx=40NF_x=40N acting on object of mass m2m_2 and also both masses are connected by string.

Now, lets draw the Free Body Diagram(FBD) for the system.


i). Now applying Newton's second law we get

Fnet=mtotalanet    (Fxfk)=(m1+m2)anet    405=(2+8)anet    anet=3.5m/s2F_{net}=m_{total}\cdot a_{net}\\ \implies (F_x-f_k)=(m_1+m_2)a_{net}\\ \implies 40-5=(2+8)a_{net}\implies a_{net}=3.5m/s^2

Thus, acceleration of each mass is 3.5m/s23.5m/s^2 .


ii). Let TT denotes the tension force.

Now,consider the FBD object of mass m1m_1


Clearly, from the above diagram we get,

Tfk=m1anet    T=fk+m1anet    T=5+2×3.5=12NT-f_k=m_1a_{net}\\ \implies T=f_k+m_1a_{net}\implies T=5+2\times3.5=12N

Thus, Tension on the string is 12N12N .


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