Two blocks A and B, attached to each other by a massless spring, are kept on a rough horizontal surface (μ=0.1) and pulled by a force F=200 N as shown in the figure. If at some instant, the 10 kg mass has acceleration of 12 m / s 2, what is the acceleration of 20 kg mass (in m / s 2 )?

Question

Two blocks A and B, attached to each other by a massless spring, are kept on a rough horizontal surface (μ=0.1) and pulled by a force F=200 N as shown in the figure. If at some instant, the 10 kg mass has acceleration of 12 m / s 2, what is the acceleration of 20 kg mass (in m / s 2 )? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/99cf4927fb6911b8052ab8b6b7d7db3c-.png" /> N1=100 k x-μ N1=10 × 12 k x-10=120 k x=130 ...(i) N2=200 200-k x-μ N2=20 a 200-130-(0.1 × 200)=20 a 70-20=20 a a=(50/20)=2.5 m / s -2

Q. Two blocks A and B, attached to each other by a massless spring, are kept on a rough horizontal surface (μ=0.1) and pulled by a force F=200N as shown in the figure. If at some instant, the 10kg mass has acceleration of 12m/s2, what is the acceleration of 20kg mass (in m/s2 )?

N1​=100 kx−μN1​=10×12 kx−10=120 kx=130 ...(i) N2​=200 200−kx−μN2​=20a 200−130−(0.1×200)=20a 70−20=20a a=2050​=2.5m/s−2

$N_{1} = 100$
$k x - μ N_{1} = 10 \times 12$
$k x - 10 = 120$
$k x = 130$ ...(i)
$N_{2} = 200$
$200 - k x - μ N_{2} = 20 a$
$200 - 130 - (0.1 \times 200) = 20 a$
$70 - 20 = 20 a$
$a = \frac{50}{20} = 2.5 m / s^{- 2}$