Question

A block of mass 15.0 kg slides from rest down a frictionless 34.0° incline and is stopped by a strong spring with k = 2.90 ✕ 104 N/m. The block slides 3.00 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?

Answer 1

Answer:

Spring is compressed by 0.158 m  

Explanation:

We have given mass of the block m = 15 kg

Angle of inclination $\Theta =34^{\circ}$

It is given that block slides 3 m from point of release

So height $h=3cos\Theta =3\times cos34^{\circ}=2.487m$

Potential energy will be equal to $U=mgh=15\times 9.8\times 2.487=365.60J$

Spring constant $k=2.90\times 10^{4}N/m$

This potential energy will be equal to energy stored in spring

So $\frac{1}{2}kx^2=365.60$

$\frac{1}{2}\times 2.9\times 10^4\times x^2=365.60$

$x=0.158m$

So spring is compressed by 0.158 m