Question

In the figure, a block of mass m = 15 kg is released from rest on a frictionless incline of angle θ = 29°. Below the block is a spring that can be compressed 2.1 cm by a force of 360 N. The block momentarily stops when it compresses the spring by 5.3 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

Answer 1

here,

the mass , m = 15 kg

theta = 29 degree

x = 2.1 cm = 0.021 m

let the spring constant be K

force = K * x

360 = K * 0.021

K = 1.71 * 10^4 N/m

a)

x2 = 5.3 cm = 0.053 m

let the distance traveled distance before stopping be s

using conservation of energy

0.5 * K * x2^2 = m * g * s

0.5 * 1.71 * 10^4 * 0.053^2 = 15 * 9.81 * s

s = 0.164 m = 16.4 cm

b)

let the speed of block just before it touchs the spring be v

using conservation of energy

m * g * (s - x2) = 0.5 * m * v^2

9.81 * ( 0.164 - 0.053) = 0.5 * v^2

v = 1.47 m/s

the speed of the block is 1.47 m/s