A weight of 290 N and another of 200 N are suspended by a rope on either side of a frictionless pulley. The acceleration of each weight is

Question

A weight of 290 N and another of 200 N are suspended by a rope on either side of a frictionless pulley. The acceleration of each weight is (A) 1.5 m text/s2  (B) 1.8 m text/s2  (C) 2.2 m text/s2  (D) 2.5 m text/s2

Tardigrade Admin · Accepted Answer

Acceleration, a=( (m2-m1/m2+m1) )g ⇒ a=( (W2-W1/W2+W1) )g =( (290-200/290+200) )× 9.8 ⇒ a=(90× 9.8/490) ⇒ a=1.8 m text/s2

Q. A weight of 290 N and another of 200 N are suspended by a rope on either side of a frictionless pulley. The acceleration of each weight is

1.5 $m / s^{2}$

1.8 $m / s^{2}$

2.2 $m / s^{2}$

2.5 $m / s^{2}$

Acceleration, $a = (\frac{m _{2} - m _{1}}{m _{2} + m _{1}}) g$ $\Rightarrow$ $a = (\frac{W _{2} - W _{1}}{W _{2} + W _{1}}) g$ $= (\frac{290 - 200}{290 + 200}) \times 9.8$ $\Rightarrow$ $a = \frac{90 \times 9.8}{490}$ $\Rightarrow$ $a = 1.8 m / s^{2}$

Q. A weight of 290 N and another of 200 N are suspended by a rope on either side of a frictionless pulley. The acceleration of each weight is

1.5 m/s2

1.8 m/s2

2.2 m/s2

2.5 m/s2

Acceleration, a=(m2​+m1​m2​−m1​​)g ⇒ a=(W2​+W1​W2​−W1​​)g =(290+200290−200​)×9.8 ⇒ a=49090×9.8​ ⇒ a=1.8m/s2

1.5 $m / s^{2}$

1.8 $m / s^{2}$

2.2 $m / s^{2}$

2.5 $m / s^{2}$

Acceleration, $a = (\frac{m _{2} - m _{1}}{m _{2} + m _{1}}) g$ $\Rightarrow$ $a = (\frac{W _{2} - W _{1}}{W _{2} + W _{1}}) g$ $= (\frac{290 - 200}{290 + 200}) \times 9.8$ $\Rightarrow$ $a = \frac{90 \times 9.8}{490}$ $\Rightarrow$ $a = 1.8 m / s^{2}$