In the figure below, a block of mass m = 3.20 kg slides from rest a distance d d

Question

In the figure below, a block of mass m = 3.20 kg slides from rest a distance d down a frictionless incline at angle = 30.0° where it runs into a spring of spring constant 435 N/m. When the block momentarily stops, it has compressed the spring by 19.5 cm.

(a) What is the distance d?


(b) What is the distance between the point of first contact and the point where the block's speed is greatest?

In the figure below, a block of mass m = 3.20 kg slides from rest a distance d down a frictionless incline at angle I = 30.0 A degree where it runs into a spring of spring constant 435 N/m. When the block momentarily stops, it has compressed the spring by 19.5 cm. (a) What is the distance d? (b) What is the distance between the point of first contact and the point where the block's speed is greatest?

Explanation / Answer

m = 3.2 kg, angle A = 30 deg, x = 19.5 cm =0.195 m

spring constant k = 435 N/m .

What is the value of d?

initial energy = mg(d + x)sinA

final energy = kx^2/2

so mg(d + x)sinA = kx^2/2

d = kx^2/(2mgsinA) - x = 0.332 m = 33.2 cm

b)

What is the distance between the point of first contact and the point where the block's speed is greatest?

speed is greatest ==> acceleration = 0 ==> net force = 0

mgsinA = kx'

Kx' = 3.2*9.81*sin30 = 15.696

x' = 15.696/435 = 0.036 m = 3.6 cm

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