Question

A block of mass m=0.4 kg starts from rest and slides down a frictionless ramp, before hitting a spring of spring constant k = 300 N/m placed on a horizontal surface. When the block is at the height of 1.5m above the ground, its velocity is 2m/s. What is the maximum compression of the spring? Take g=9.8 m/s^2.

Answer 1

Answer:

Maximum spring compression is 0.21 m

Solution:

As per the question:

Mass of the block, M = 0.4 kg

Spring constant, k = 300 N/m

Height of the block, h = 1.5 m

Velocity of the block, v = 2 m/s

Now,

By conservation of energy:

$PE + KE = E_{spring}$                   (1)

where

PE = Potential energy = Mgh

KE = Kinetic Energy = $\frac{1}{2}Mv^{2}$

$E_{spring} = \frac{1}{2}kx^{2}$ = energy in the spring

Now, using eqn (1) to calculate maximum compression 'x' of the spring:

$Mgh + \frac{1}{2}Mv^{2} = \frac{1}{2}kx^{2}$

$0.4\times 9.8\times 1.5 + \frac{1}{2}\times 0.4\times (2)^{2} = \frac{1}{2}\times 300x^{2}$

$x^{2} = 0.0445$

x = 0.21 m