Answer to Question #258037 in Mechanics | Relativity for Bak

Question

A 200-g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.08 to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops (a) if the ramp exerts no friction force on the block and (b) if the coefficient of kinetic friction is 0.400.

Answer 1

m=200 g = 0.2 kg

k= 1.4 kN/m $= 1.4 \times 10^3 \; N/m$

x= 10 cm = 0.1 m

θ = 60°

(a) Let the distance moved by the block along frictionless ramp as d

Then the height rised by the block is

h =d sinθ

From the law of conservation of mechanical energy

$K_i +U_{gi} +U_{si} = F_f +U_{gf} +U_{sf} \\ 0 + 0 + \frac{1}{2}kx^2_i = 0 +mg(dsin θ) + 0 \\ \frac{1}{2}kx^2 = mg(dsin θ) \\ d = \frac{kx^2_i}{2(mgsin θ)} \\ = \frac{1.4 \times 10^3 \times 0.1}{2 \times 0.2 \times 9.8 \times sin 60°} \\ = 4.12 \;m$

(b)

Let the distance moved by the block along the ramp having a coefficient of friction $\mu_k=0.4$ is d.

The change in energy equals the amount of work done on an object according to the work-energy theorem. So we have

$ΔE = W_f$

(work done by the frictional force)

Now we have to subtracting the work done from the potential energy of the spring

From law of conservation of mechanical energy

$K_i +U_{gi} +U_{si} -W_f = K_f +U_{gf} +U_{sf} \\ 0 + 0 + \frac{1}{2}kx^2_i -(\mu_k mg cos θ)d = 0 +mg(dsin θ) + 0 \\ \frac{1}{2}kx^2_i -(\mu_k mg cos θ)d = mg(d sin θ) \\ \frac{1}{2}kx^2_i = (\mu_k mg cos θ)d +mg(d sin θ) \\ d = \frac{kx^2_i}{2(mg sin θ + \mu_k mg cos θ)} \\ d = \frac{kx^2_i}{2mg(sin θ + \mu_k cos θ)} \\ = \frac{1.4 \times 10^3 \times 0.1}{2 \times 0.2 \times 9.8 (sin60° + 0.4 \times cos60°) } \\ = 3.35 \;m$