# A 10kg block on a table conected to a 78kg mass (find acceleration)

by Zsmitty3
Tags: 10kg, 78kg, acceleration, block, conected, mass, table
 P: 46 1.A 10 kg block (A) on a table is connected by a string to a 78-kg mass (B), which is hanging over the edge of the table. Assuming that frictional forces may be neglected, what is the magnitude of acceleration of the 10kg black when the other block is released? 2.So the acceleration of the block (A) is going to be the same as the acceleration of the Mass (B) in the y direction? Where the block =A and the Mass =B 3. block x: T=m(A)a Mass y: T-m(B)g= -m(B)a Substitute in T in the Mass y equation: ma-mg=-ma Therefore: a= (m(B)g) / (m(B)-m(A)) (7.8*9.8)/(78+10)=8.7? Seems a little quick to be right?
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Thanks
P: 10,539
 Quote by Zsmitty3 1.A 10 kg block (A) on a table is connected by a string to a 78-kg mass (B), which is hanging over the edge of the table. Assuming that frictional forces may be neglected, what is the magnitude of acceleration of the 10kg black when the other block is released? 2.So the acceleration of the block (A) is going to be the same as the acceleration of the Mass (B) in the y direction? Where the block =A and the Mass =B 3. block x: T=m(A)a Mass y: T-m(B)g= -m(B)a
Correct so far.

 Quote by Zsmitty3 Substitute in T in the Mass y equation: ma-mg=-ma????? Therefore: a= (m(B)g) / (m(B)-m(A))
That is wrong. Check.

ehild

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