Two objects are connected by a string of length 4.8 m and negligible mass. The 10 kg block is placed on a smooth table top(3.6 m from end of table),and the 7kg block hangs over the edge of the table. The 7 kg block is then released from rest at a distance of 1.2 m above the ﬂoor at time t = 0.The acceleration of gravity is 9.8 m/s^2. The acceleration of the 7kg block is 4.04 m/s^2. The 7 kg block strikes the ﬂoor and doesnot bounce.How long dies ittakethe7kgblocktostrikethe ﬂoor? Answer in units of s

Thank you for the opportunity to help you with your question! the solution is as such:

The acceleration acting on the m= 7 kg block is (9.8 + 4.04). soit os the total force on it is f= ma

it is f+ 7*(4.04+ 9.8) = 96.88N

now this force acts as tension through the string to the m' 10 kg block , so its acceleration will be a'

now f= m' *a'

so a' = 96.88/ 10 = 9.688 m/s

this acceleration produces a reversing acceleration to the 10 kg block on the table. so the net acceleration on the 7 kg block is 13.84- 9.688 = 4.15 m/s^2

now since the 7 kg block starts from rest, so u=0.

applying, the motion equation, s= u*t + 0.5 *a*(t) ^2

and 1.2 = 0*t + 0.5 *4.15*(t)^2

we get , t= 0.76s

so the time taken is 0.76 seconds

Please let me know if you need any clarification. I'm always happy to answer your questions.awaiting a positive feedback. thank you.

the same tension acts in both the parts of the string.

the tension is T=[(m1*m2) / ( m1 + m2)] *g

so the tension is 4.117 N

where m1 = 7kg and m2 = 10 kg and g = 9.8

now the acceleration of the falling block is given by a= [m1/( m1+m2)] * g

so a = [7/(7+10)] *9.8

a= 4.035m/ s^2

now

now since the 7 kg block starts from rest, so u=0.

applying, the motion equation, s= u*t + 0.5 *a*(t) ^2

and 1.2 = 0*t + 0.5 *4.035*(t)^2

we get , t= 0.77s

so the time taken is 0.77 seconds. Please
let me know if you need any clarification. I'm always happy to answer
your questions.awaiting a positive feedback. thank you.

Jun 21st, 2015

did u get the final answer?

Jun 23rd, 2015

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