Two masses of 5 kg and 10 kg are connected to a pulley as shown. What will be the acceleration of the system? (g = acceleration due to gravity)

Question

Two masses of 5 kg and 10 kg are connected to a pulley as shown. What will be the acceleration of the system? (g = acceleration due to gravity) (A) (g/4) (B) (g/3) (C) (g/2) (D) g

Tardigrade Admin · Accepted Answer

According to Newton's law, ∑ F1= T -5 g = m 1 a ... (i) ∑ F2=10 g - T = m 2 a ...(ii) Adding, (i) and (ii) T -5 g =5 × a (10 g - T +10 × a/5 g =15 a) a = g / 3

Q. Two masses of $5 k g$ and $10 k g$ are connected to a pulley as shown. What will be the acceleration of the system? ( $g$ = acceleration due to gravity)

$\frac{g}{4}$

$\frac{g}{3}$

$\frac{g}{2}$

$g$

According to Newton's law,
$\sum F_{1} = T - 5 g = m_{1} a ... (i)$
$\sum F_{2} = 10 g - T = m_{2} a ... (ii)$
Adding, (i) and (ii)
$T - 5 g = 5 \times a$
$\frac{10 g - T + 10 \times a}{5 g = 15 a}$
$a = g /3$

Q. Two masses of 5kg and 10kg are connected to a pulley as shown. What will be the acceleration of the system? (g = acceleration due to gravity)

4g​

3g​

2g​

g

According to Newton's law, ∑F1​=T−5g=m1​a...(i) ∑F2​=10g−T=m2​a...(ii) Adding, (i) and (ii) T−5g=5×a 5g=15a10g−T+10×a​ a=g/3

Q. Two masses of $5 k g$ and $10 k g$ are connected to a pulley as shown. What will be the acceleration of the system? ( $g$ = acceleration due to gravity)

$\frac{g}{4}$

$\frac{g}{3}$

$\frac{g}{2}$

$g$

According to Newton's law,
$\sum F_{1} = T - 5 g = m_{1} a ... (i)$
$\sum F_{2} = 10 g - T = m_{2} a ... (ii)$
Adding, (i) and (ii)
$T - 5 g = 5 \times a$
$\frac{10 g - T + 10 \times a}{5 g = 15 a}$
$a = g /3$