Consider the system shown in figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (take g = 10 m s-2)

Question

Consider the system shown in figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (take g = 10 m s-2) (A) 0 N (B) 1 N (C) 2 N (D) 5 N

Tardigrade Admin · Accepted Answer

If a is acceleration of m2, then from fig. m2g-T=m2a ...(i) and T =m1a=1a ...(ii) From (i), m2 g - T = m2a 1 g - m1a = m2a 1 g = (m1 + m2)a = (1 + 1)a a=(g/2)=(10/2)=5m s-2 ∴ T=m1a=1 × 5 = 5 N

Q. Consider the system shown in figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (take g = 10 m s^-2)

0 N

1 N

2 N

5 N

If a is acceleration of m₂, then from fig.
$m_{2} g - T = m_{2} a ... (i)$
and T $= m_{1} a = 1 a ... (ii)$
From $(i), m_{2} g - T = m_{2} a$
1 g - m $_{1}$ a = m $_{2}$ a
1 g = $(m_{1} + m_{2}) a = (1 + 1) a$
$a = \frac{g}{2} = \frac{10}{2} = 5 m s^{- 2}$
$∴ T = m_{1} a = 1 \times$ 5 = 5 N

Q. Consider the system shown in figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (take g = 10 m s-2)

0 N

1 N

2 N

5 N

If a is acceleration of m2, then from fig. m2​g−T=m2​a...(i) and T =m1​a=1a...(ii) From (i),m2​g−T=m2​a 1 g - m1​a = m2​a 1 g = (m1​+m2​)a=(1+1)a a=2g​=210​=5ms−2 ∴T=m1​a=1× 5 = 5 N

Q. Consider the system shown in figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (take g = 10 m s^-2)

If a is acceleration of m₂, then from fig.
$m_{2} g - T = m_{2} a ... (i)$
and T $= m_{1} a = 1 a ... (ii)$
From $(i), m_{2} g - T = m_{2} a$
1 g - m $_{1}$ a = m $_{2}$ a
1 g = $(m_{1} + m_{2}) a = (1 + 1) a$
$a = \frac{g}{2} = \frac{10}{2} = 5 m s^{- 2}$
$∴ T = m_{1} a = 1 \times$ 5 = 5 N