A 242-g block is pressed against a spring of force constant 1.62 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions. (a) if the ramp exerts no friction force on the block (b) if the coefficient of kinetic friction is 0.440

A) d = 3.94m

B) d' = 3.14m

Explanation:

We are given;

Mass of block; m = 242g = 0.242 kg

Spring constant; k = 1.62 kN/m = 1.62 x 10³ N/m

Extension of spring; x = 10 cm = 0.1m

Coefficient of friction; µ = 0.44

Angle of inclination; θ = 60°

A) Equating the energy at the bottom of the ramp to the energy at a distance d up the ramp (at a height h in the vertical direction) we find;

W = (1/2) kx² = (1/2) •1.62 x 10³•0.1² = 8.1 J

W = mgh = 8.1

Where, h = d•sinθ

Thus, 8.1 = 0.242 x 9.8 x d sin 60

8.1 = 2.05387d

d = 8.1/2.05387

d = 3.94m

B) Here we equate the initial energy, still 8.1J, at the bottom of the ramp to the total energy at a distance d' up the ramp (at a height h') which is the sum of the gravitational potential energy (at h') and the energy losses due to

friction. The energy losses due to friction are calculated from the work

done by friction which is the force of friction times the displacement of

the object. The force of friction is simply the product of the normal force and the coefficient of kinetic friction. The normal force is found through a free body diagram and has magnitude F_n = mg cos θ. Putting this all together, we have;

8.1 = mgh' + ΔE

Thus,

8.1 = mgd'•sin θ + µmgd'•cos θ

So, d' = 8.1/[mgd'•sin θ + µmgd'•cos θ]

Plugging in relevant values;

8.1/[0.242x9.8 (d'sin60 + (0.44 x d' x cos60))

8.1 = 2.3716d' (0.866 + 0.22)

8.1 = d' (2.3716 x 1.086)

d' = 8.1/2.576 = 3.14m

a) = 3.94 m

b) = 3.15 m

Explanation:

Given

Mass of the block, m = 242 g

Force constant, k = 1.62 kN/m

Compression of the spring, x = 10 cm

Angle of inclination = 60°

a) if we equate the energy at the bottom of the ramp to the energy at a distance d up the ramp, we have

1/2kx² = mgh where, h = dsinΦ

1/2kx² = mgdsinΦ

1/2 * 1.62*10^3 * 0.1² = 0.242 * 9.8 * dsin 60

1/2 * 16.2 = 2.3716 * d sin 60

d sin 60 = 8.1 / 2.3716

0.866 d = 3.415

d = 3.415 / 0.866

d = 3.94 m

b) net force on the block = mgd sin 60 + µ mgd cos 60

8.1 = d[mg sin 60 + µ mg cos 60]

8.1 = d [0.242 * 9.8 * 0.866 + 0.44 * 0.242 * 9.8 * 0.5]

8.1 = d (2.05 + 0.52)

8.1 = 2.57 d

d = 8.1 / 2.57

d = 3.15 m