Question

Two bodies whose masses are m1 = 50 kg and my
= 150 kg are tied by a light string and are placed on
a frictionless horizontal surface. When my is pulled
by a force F, an acceleration of 5 ms 2 is produced
in both the bodies. Calculate the value of F. What
is the tension in the string 1 ?​

Answer 1

Answer:  Force of both bodies are 250 N and 750 N respectively.

Tension in string is 2220 N.

Explanation:

Since we have given that

Mass of first body = 50 kg

Mass of second body = 150 kg

Acceleration in two bodies = 5m/s²

So, Force exerted in first body is given by

$F=ma\\\\F=50\times 5\\\\F=250\ N$

Similarly,

Force exerted in second body is given by

$F=ma\\\\F=150\times 5\\\\F=750\ N$

Hence, Force of both bodies are 250 N and 750 N respectively.

Tension in the string 1 is given by

$mg+ma\\\\=m(g+a)\\\\=150(9.8+5)\\\\=150\times 14.8\\\\=2220\ N$

Hence, the tension in string is 2220 N.

Answer 2

Answer:

Answer: Force of both bodies are 250 N and 750 N respectively.

Tension in string is 2220 N.

Explanation:

Since we have given that

Mass of first body = 50 kg

Mass of second body = 150 kg

Acceleration in two bodies = 5m/s²

So, Force exerted in first body is given by

\begin{gathered}F=ma\\\\F=50\times 5\\\\F=250\ N\end{gathered}

F=ma

F=50×5

F=250 N

Similarly,

Force exerted in second body is given by

\begin{gathered}F=ma\\\\F=150\times 5\\\\F=750\ N\end{gathered}

F=ma

F=150×5

F=750 N

Hence, Force of both bodies are 250 N and 750 N respectively.

Tension in the string 1 is given by

\begin{gathered}mg+ma\\\\=m(g+a)\\\\=150(9.8+5)\\\\=150\times 14.8\\\\=2220\ N\end{gathered}

mg+ma

=m(g+a)

=150(9.8+5)

=150×14.8

=2220 N

Hence, the tension in string is 2220 N.