Question

1 a block of mass 1.7 kg lies on a frictionless surface. a man slides the block against a spring, compressing it 15 cm. when the man lets go of the spring, the block moves at 5 m/s. what is the spring constant of the spring?

Explanation:

Step1: Identify the physics concept

This problem involves conservation of mechanical energy (kinetic and elastic potential energy) on a frictionless surface. The formula for elastic potential energy is \( U = \frac{1}{2}kx^2 \) and kinetic energy is \( K = \frac{1}{2}mv^2 \). At maximum compression, the kinetic energy of the block is converted into elastic potential energy of the spring (assuming initial potential energy of the spring is zero and we consider the point where the man lets go as the initial state with only kinetic energy for the block).

Step2: Set up the energy conservation equation

Initial kinetic energy of the block (\( K_i \)) equals the elastic potential energy of the spring at maximum compression (\( U_f \)):
\[
\frac{1}{2}mv^2=\frac{1}{2}kx^2
\]
We can cancel out the \(\frac{1}{2}\) from both sides:
\[
mv^2 = kx^2
\]

Step3: Solve for the spring constant \( k \)

Rearrange the formula to solve for \( k \):
\[
k=\frac{mv^2}{x^2}
\]
Given: \( m = 1.7\space kg \), \( v = 5\space m/s \), \( x = 0.15\space m \) (assuming the "15 m" is a typo and should be 0.15 m, as 15 m compression for a spring with a block moving at 5 m/s and mass 1.7 kg would give an unrealistically small spring constant, so likely 0.15 m).

Substitute the values:
\[
k=\frac{1.7\times(5)^2}{(0.15)^2}
\]
First, calculate the numerator: \( 1.7\times25 = 42.5 \)
Then, calculate the denominator: \( 0.0225 \)
Now, divide: \( k=\frac{42.5}{0.0225}\approx1888.89\space N/m \)

If we assume the compression is 0.15 m (correcting the likely typo from 15 m to 0.15 m, since 15 m would be impractical). If the compression is indeed 15 m (which is very large), then:

\[
k=\frac{1.7\times25}{225}=\frac{42.5}{225}\approx0.1889\space N/m
\]
But this is highly unlikely, so we proceed with \( x = 0.15\space m \).

Answer:

If \( x = 0.15\space m \), the spring constant \( k\approx1889\space N/m \) (or \( 0.189\space N/m \) if \( x = 15\space m \), but the former is more physically reasonable).