Question

A pulley system is connected as shown in the figure. If the spring is elongated by a distance of $0.02\text{m}$, then what is the force constant of the spring? (Assume that the spring does not affect the overall acceleration of the bodies.)

• A. $200\text{N/m}$
• B. $8000\text{N/m}$
• C. $2000\text{N/m}$
• D. $800\text{N/m}$

Answer 1

The correct option is C $2000\text{N/m}$

FBD of $5\text{kg}$ block and the whole system :



The mass on the right side of the pulley is higher. So, the direction of the acceleration will be as shown in the figure.

The acceleration of the system is given by,
$a=\frac{mg+m_{1}g-mg}{(m+m+m_1)}=\frac{5\times 10}{25}\Rightarrow a=2m/s^2$

From the free body diagram of $5\text{kg}$, we have the equation,
$50-T=5\times 2\Rightarrow T=50-10=40N$

The tension in the spring is also $T$
$T=kx\Rightarrow 40=k\times 0.02\Rightarrow k=\frac{40}{0.02}=2000N/m$