A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

x = 0.056 m

ΔKE = 0.489 J

Explanation:

Given that

Angle, θ = 38°

Length, L = 1.7 m

Mass, m = 0.09 kg

Spring constant, K = 590 N/m

If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE

This is mathematically written as

1/2kx² = mgH

The height, H we can get by using the relation

H = L.Sinθ

H = 1.7 * Sin 38

H = 1.7 * 0.6157

H = 1.047 m

Next, we use the Work-Energy theorem

1/2kx² = mgH

1/2 * 590 * x² = 0.09 * 9.8 * 1.047

295 * x² = 0.9234

x² = 0.9235 / 295

x² = 0.00313

x = √0.00313

x = 0.056 m

If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8

Then we use the formula

ΔKE = mg(H - H1)

ΔKE = mg(xsinθ - x2.sinθ)

Where, x = 1.7 , x2 = 0.8

ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)

ΔKE = 0.882(1.047 - 0.493)

ΔKE = 0.882 * 0.554

ΔKE = 0.489 J

. west

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray