Question

A 1.50-kg block is on a frictionless, 30 degrees inclined plane. The block is attached to a spring (k = 40.0N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.40m/s. How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)

y = _____ m

Please show steps I am confused.

Answer 1

Solution: -
Let
M1 = block's mass,
M2 = suspended mass
theta = angle of incline

M1 = 1.50 kg
M2 = 60.0 g = 60.0 * 10^-3 kg = 0.06 kg
theta = 30 deg

Forces on M2 are: -
1. Weight M2 * g downward
2. Tension T in the string upward
Initially M2 is in equilibrium, therefore net force on it = 0
Therefore, T = M2 * g
Or T = 0.06 * 9.8
Or T = 0.588 N------------------------(1)

Let x1 = length by which the spring is initially compressed. Forces on the block parallel to the incline are: -
1. Force k*x1 = 40 x1 by spring up the incline
2. Component of weight = M1*g * sin(theta)
= 1.50 * 9.8 * sin(30 deg) = 7.35 N down the incline
3. Tension T up the incline

The block is initially in equilibrium. Therefore, net force in any direction = 0
Therefore, force up the incline = force down the incline
Or 40 x1 + T = 7.35
Or 40 x1 = 7.35 - T
Using equation (1) in the above: -
Or 40 x1 = 7.35 - 0.392
Or 40 x1 = 6.96
Or x1 = 6.96/40
Or x1 = 0.174 m -------------------------(2)

Consider block+suspended mass+spring+string as our system.
Let,
Speed given to suspended mass = 1.40 m/s

Initial kinetic energy of the system = 1/2*M2*1.40^2
= 1/2 * 0.04 * 1.4^2
= 0.0392 J------------------------(3)
Let the mass drop to height h. After dropping to this height, it comes to rest. Therefore, final kinetic energy of the system = 0
Therefore, change in kinetic energy = final kinetic energy - initial kinetic energy
Or change in kinetic energy = 0 - 0.0392 J
Or change in kinetic energy = - 0.0392 J ------------------(4)

Let U1s = initial potential energy of the spring
U1s = 1/2 * k * x1^2
Using value of x1 from equation (2) and the given value of k,
U1s = 1/2 * 40 * 0.174^2
Or U1s = 0.605 J----------------------(5)

Let L = natural length of the spring
Then initial length of the spring = L - x1 because it is compressed by length x1
As the block drops to height h, the length of the spring becomes L - x1 + h
Therefore, change in length compared to the natural length
= (L - x1 + h) - L
= h - x1 = h - 0.174 (taking x1 from eq (2))
Let U2s = final potential energy of the spring
U2s = 1/2 * k * (h - 0.174)^2
O U2s = 1/2 * 40 * (h - 0.174)^2
Or U2s = 20 (h - 0.174)^2 ----------------(6)

Suspended mass drops by height h.
Therefore its gravitational potential energy(GPE) decreases by M2*gh
= 0.06 * 9.8 * h = 0.588 h
Therefore,
change in GPE of suspended mass
= -0.588 h-----------------(7)
(Minus sign is used in the above because GPE decreases.)

The block moves up the incline by distance h. Therefore, vertically upward distance by the block = h * sin(theta)
= h * sin(30 deg) = 0.5 h
Therefore, its GPE increases by M1*g*0.5 h = 1.50 * 9.8 * 0.5 h
= 7.35 h
Therefore,
change in GPE of the block
= 7.35 h----------------------(8)

There is no friction. Therefore, kinetic energy + potential energy is conserved.
Or change in kinetic energy + change in potential energy
= 0---------------------------(9)

Change in PE = change in spring's PE + change in suspended mass's GPE + change in block's GPE

Or Change in PE = U2s - U1s + change in suspended mass's GPE + change in block's GPE
Using equations (5), (6), (7), (8) in the above: -
Change in PE
= 20(h - 0.174)^2 - 0.605 - 0.588 h + 7.35 h---------------(10)

Using equations (4) and (10) in (9) : -

-0.0392 + 20(h - 0.174^2 - 0.605 - 0.588 h + 7.35 h = 0
Or 20(h - 0.174)^2 - 0.644 + 6.762 h = 0
Or 20(h^2 - 0.348 h + 0.0302) - 0.644 + 6.76 h = 0
Or 20 h^2 - 6.96 h + 0.604 - 0.644 + 6.76 h = 0
Or 20 h^2 - 0.2 h - 0.04 = 0
h = 0.05 Or h = -0.04

h cannot be negative.. so,
Or h = 0.05 m