1.

Consider the case of two bodies of masses m_(1) and m_(2) which are connected by light inextensible string passing over a light smooth pulley as shown in the figure. The expression for acceleration of the system and tension of the string are expressed below under different situations: (i) when m_(1) gt m_(2). In this case a=((m_(1)-m_(2))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g (ii) when m_(2) gt m_(1). In this case a=((m_(2)-m_(1))/(m_(1)+m_(2)))g and T=((2m_(1)m_(2))/(m_(1)+m_(2)))g (iii) When m_(1)=m_(2)=m. In this case a = 0, T = mg If the pulley is pulled upward with acceleration equal to the acceleration due to gravity, what will be the tension in the string?

Answer»

75N
150N
300N
500N

Solution :`2m_(1)g-T.=m_(1)a. "" ...(1) `
`T.-2m_(2)g=m_(2)a. ""...(2)`
On SOLVING (1) and (2)
`a.=(g)/(2)=5m//s^(2)`
`T.=150N`


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