Question

A 225-g block is pressed against a spring of force constant 1.26 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
m

(b) if the coefficient of kinetic friction is 0.369
m

Answer 1

As we know that;

a)

= Initial energy = Final energy
= (1/2)kx^2 = mgL(sin 60)
where,
= k = spring constant = 1260N/m
= x = initial spring deflection = 10 cm = 0.10 m
= m = mass of the block = 225 g = 0.225 kg.
= g = acceleration due to gravity = 9.8 m/sec^2
= L = distance along incline travelled by block

So,
Substituting appropriate values,
= (1/2)(1260)(0.10)^2 = 0.225(9.8)(L)(sin 60)
Solving for L,
= L = 3.299 meters

Then,

b)

Working equation will be,
= (1/2)kx^2 - µ(mg)(cos 60)L = mgL(sin 60)
where,
= µ = coefficient of friction = 0.369
All the rest of terms have been previously defined,
So, Substituting values,
= (1/2)(1260)(0.10)^2 - 0.369(0.225)(9.8)(cos 60)(L) = 0.225(9.8)(L)(sin 60)
= 6.3 - 0.406L = 1.909 L
= 1.909L + 0.406L = 6.3
So,

Solving for L,
= L = 2.72 m