Two blocks of masses of 40 kg and 30 kg are connected by a weightless string passing over a frictionless pulley as shown in the figure.

Options

A. `0.7"ms"^(-2)`
B. `0.8"ms"^(-2)`
C. `0.6"ms"^(-2)`
D. `0.5"ms"^(-2)`

Answer

A

Explanation

Hence, `m_(1)=40g`
`m_(2)=30kg, theta=30^(@)`
Let T be the tension in the string and a be the acceleration of the system. Their equations of motion are `m_(1)sin 30^(@)-T=m_(1)a…(i)`
`T-mg sin 30^(@)=m_(2)a…(ii)`
Adding (i) and (ii), we get
`(m_(1)+m_(2))a=(m_(1)-m_(2))g sin 30^(@)`
Substituting the given values, we get
`(40+30)a=(40-30)xx9.8xx(1)/(2)=49`
`a=(49)/(70)=0.7ms^(-2)`