Question #249846

two blocks of masses m1= 4.5 kg and m2= 6.5 kg resting on a frictionless surface connected by a light inextensible cord as shown in figure 3. a horizontal force F of 33 N directed to the right is applied to the block with m1 as shown. Find (a) the acceleration of the masses and (b) the tension T in the cord between the two blocks


1
Expert's answer
2021-10-12T10:22:39-0400



Newton's Second Law of Motion


FT=m1aF-T=m_1a

T=m2aT=m_2a

(a)


(m1+m2)a=F(m_1+m_2)a=F

a=Fm1+m2a=\dfrac{F}{m_1+m_2}

a=33 N4.5 kg+6.5 kg=3 m/s2a=\dfrac{33\ N}{4.5\ kg+6.5\ kg}=3\ m/s^2

(b)


T=m2aT=m_2a

T=6.5 kg(3 m/s2)=19.5 NT=6.5\ kg(3\ m/s^2)=19.5\ N


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Comments

Simon
11.11.21, 06:14

Thank you so much. you are a life saver!

Roan
13.10.21, 05:25

Thank you so much po you did great and very helpful,godbless

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