Awooden block with mass 1.45 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 29.0 degrees (point a). when the spring is released, it projects the block up the incline. at point b, a distance of 4.60 m up the incline from a, the block is moving up the incline at a speed of 5.10 m/s and is no longer in contact with the spring. the coefficient of kinetic friction between the block and incline is μk=0.45. the mass of the spring is negligible. a.calculate the amount of potential energy that was initially stored in the spring. take free fall acceleration to be 9.80 m/s2 .

76.3 J

Explanation:

I'm assuming the distance of 4.60 m is along the incline, not the vertical distance from the bottom.  I'll call this distance d, so h = d sin θ.

Initial energy = final energy

Energy in spring = gravitational energy + kinetic energy + work by friction

E = mgh + 1/2 mv² + Fd

We need to find the force of friction.  To do that, draw a free body diagram.

Normal to the incline, we have the normal force pointing up and the normal component of weight (mg cos θ).

Sum of the forces in the normal direction:

∑F = ma

N - mg cos θ = 0

N = mg cos θ

Friction is defined as:

F = Nμ

Plugging in the expression for N:

F = mgμ cos θ

Substituting:

E = mgh + 1/2 mv² + (mgμ cos θ) d

E = mg (d sin θ) + 1/2 mv² + (mgμ cos θ) d

E = mgd (sin θ + μ cos θ) + 1/2 mv²

Given:

m = 1.45 kg

g = 9.90 m/s²

d = 4.60 m

θ = 29.0°

μ = 0.45

v = 5.10 m/s

Solving:

E = mgd (sin θ + μ cos θ) + 1/2 mv²

E = (1.45) (9.80) (4.60) (sin 29.0 + 0.45 cos 29.0) + 1/2 (1.45) (5.10)²

E = 76.3 J

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach