Question

suppose its initial speed of the block is 1.39 m/s but its mass can be varied what mass is required to give the maximum spring compression of 3.55 cm

Answer 1

Answer:

m = 0.623 kg

Explanation:

By the conservation of energy, we can write the following equation

$\begin{gathered} E_i=E_f \\ KE=PE \\ \frac{1}{2}mv^2=\frac{1}{2}kx^2 \end{gathered}$

Where m is the mass, v is the speed, k is the constant of the spring and x is the compression.

Solving the equation for m, we get

$\begin{gathered} 2^\cdot\frac{1}{2}mv^2=2\cdot\frac{1}{2}kx^2 \\ \\ mv^2=kx^2 \\ \\ m=\frac{kx^2}{v^2} \end{gathered}$

Now, we can replace k = 955 N/m, x = 3.5 cm = 0.0355 m, and v = 1.39 m/s to get

$\begin{gathered} m=\frac{(955\text{ N/m\rparen\lparen0.0355 m\rparen}^2}{(1.39\text{ m/s\rparen}^2} \\ \\ m=0.623\text{ kg} \end{gathered}$

Therefore, the mass required is 0.623 kg