Question

Two Masses Of 10 Kg And 20 Kg Respectively Are Tied Together By A Massless Spring. A Force Of 200 N Is Applied On The 20 Kg Mass. At The Instant Shown In Figure, The Acceleration Of 10 Kg Mass Is 12 M/S2, The Acceleration Of 20 Kg Mass Is ______

Answer 1

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(-0.5,-1)(1.7,0){10}{\qbezier(-0.5,-1)(0,0)(0,0)\qbezier(0,0)(1,2)(1.5,-1)\qbezier(1.5,-1)(1.6,-1.8)(1.4,-3)\qbezier(1.4,-3)(0.9,-3)(1.2,-1)}\qbezier(16.5,-1)(17,0)(17,0)\qbezier(-0.5,-1)(0,0)(0,0)\qbezier(17,0)(18,0)(18.5,-2)\multiput(-10.5,-2)(29,0){2}{\line(1,0){10}}\put(-20.5,-7){\framebox(10,10){\sf{10 kg}}}\put(28.5,-7){\framebox(10,10){\sf{20 kg}}}\put(38.5,-2){\vector(1,0){10}}\put(49,-3){\sf{200 N}}\put(-30,-7){\line(1,0){80}}\end{picture}$

Assume the horizontal surface to be smooth.

Free body diagram of the masses are given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(0,20){2}{\circle*{1}}\multiput(0,0)(0,20){2}{\vector(1,0){10}}\put(0,20){\vector(-1,0){10}}\put(-4,16){\sf{20\ kg}}\put(-4,3){\sf{10 kg}}\multiput(-14.5,19)(25.5,-20){2}{\sf{kx}}\put(11,19){\sf{200 N}}\multiput(-4.5,-6)(0,31){2}{\vector(1,0){10}}\put(-5,-10){ \sf{12\ m\,s^{-2}} }\put(-1,27){ \sf{a} }\end{picture}$

Here $\sf{kx}$ is the restoring force produced in the spring.

From FBD of 20 kg mass, we get,

$\longrightarrow\sf{200-kx=20a\quad\quad\dots(1)}$

And from FBD of 10 kg mass, we get,

$\longrightarrow\sf{kx=10\times12}$

$\longrightarrow\sf{kx=120\quad\quad\dots(2)}$

Adding (1) and (2),

$\longrightarrow\sf{200-kx+kx=20a+120}$

$\longrightarrow\sf{200=20a+120}$

$\longrightarrow\sf{a=\dfrac{200-120}{20}}$

$\longrightarrow\sf{\underline{\underline{a=4\ m\,s^{-2}}}}$

Answer 2

Answer:

8m/s^2

Explanation:

from fbd of 10 kg

M x A = KX

10 X 4 = Kx           (1)

fbd of 20 kg

force applied - kx = 20a                    (2)

 from equation 1 and 2

200 -40  = 20 a

then a = 8 m/s^2