Question

Calculate the acceleration of mass at the maximum elongation of spring if a constant force acts on block placed on smooth surface if the spring shown is in its natural length

• A. $\frac{2F}{m}$
• B. $\frac{F}{m}$
• C. Zero
• D. $\frac{F}{2m}$

Answer 1

Answer: (B)

$\frac{F}{m}$

Answer 2

To calculate the acceleration of the mass at the maximum elongation of the spring, we need to follow these steps:

1. Determine the maximum elongation ($x_{max}$):
   The block starts from rest at the natural length of the spring (initial position $x_i = 0$, initial velocity $v_i = 0$). At maximum elongation, the block momentarily comes to rest (final velocity $v_f = 0$). We can use the Work-Energy Theorem, which states that the net work done on the block equals its change in kinetic energy ($W_{net} = \Delta KE$).

   • Work done by the constant force $F$: $W_F = F \cdot x_{max}$
   • Work done by the spring force: The spring force is conservative, and its potential energy is $U_s = \frac{1}{2}kx^2$. The work done by the spring force is $W_s = -\Delta U_s = -(U_{s,f} - U_{s,i}) = -(\frac{1}{2}kx_{max}^2 - \frac{1}{2}k(0)^2) = -\frac{1}{2}kx_{max}^2$.

   Applying the Work-Energy Theorem:
   $W_{net} = W_F + W_s = KE_f - KE_i$
   $F \cdot x_{max} - \frac{1}{2}kx_{max}^2 = 0 - 0$
   $F \cdot x_{max} = \frac{1}{2}kx_{max}^2$

   Since $x_{max} \neq 0$ (as there is elongation), we can divide both sides by $x_{max}$:
   $F = \frac{1}{2}kx_{max}$
   From this, the maximum elongation is:
   $x_{max} = \frac{2F}{k}$

2. Calculate the acceleration at maximum elongation:
   At the point of maximum elongation ($x = x_{max}$), the forces acting on the block are:

   • The constant applied force $F$ (acting to the right).
   • The spring restoring force $F_s = kx_{max}$ (acting to the left, opposing the elongation).

   Using Newton's Second Law ($F_{net} = ma$):
   Let's take the direction of the applied force $F$ as positive.
   $F_{net} = F - F_s$
   $F_{net} = F - kx_{max}$

   Substitute the value of $x_{max}$ we found:
   $F_{net} = F - k \left(\frac{2F}{k}\right)$
   $F_{net} = F - 2F$
   $F_{net} = -F$

   Now, calculate the acceleration $a$:
   $ma = F_{net}$
   $ma = -F$
   $a = -\frac{F}{m}$

   The negative sign indicates that the acceleration is in the direction opposite to the applied force $F$ (i.e., to the left). The question asks for "the acceleration", and the options are magnitudes. Therefore, the magnitude of the acceleration is $\frac{F}{m}$.