Question

A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slope which makes an angle of 41 ∘ with respect to the horizontal. The spring is then released
A)If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
B)If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

Answer 1

Answer:

Part a)

$L = 0.68 m$

Part b)

$L = 0.63 m$

Explanation:

Part a)

As we know that there is no friction in the path

So here we can use energy conservation to find the distance moved by the mass

Initial spring energy = final gravitational potential energy

so we will have

$\frac{1}{2}kx^2 = mgL sin\theta$

$\frac{1}{2}70(0.5)^2 = 2(9.81)(L) sin41$

$8.75 = 12.87 L$

$L = 0.68 m$

Part b)

Now if spring is connected to the block then again we can use energy conservation

so we will have

$\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2$

so we will have

$\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2$

$8.75 = 6.43 + 12.87 x' + 35 x'^2$

$x' = 0.13 m$

so total distance moved upwards is

$L = 0.5 + 0.13 = 0.63 m$