Question

A 3 kg block collides with a massless spring of spring constant 98 N/m attached to a wall. The speed of the block was observed to be 1.5 m/s at the moment of collision. The acceleration of gravity is 9.8 m/s 2 . How far does the spring compress if the surface on which the mass moves is frictionless

Answer 1

The spring compresses by 22.9 cm when a 3 kg block moving at 1.5 m/s collides with it on a frictionless surface.

Step-by-step explanation:

To determine how far the spring compresses when a 3 kg block moving at 1.5 m/s collides with it, we use conservation of energy principles. Initially, the block has kinetic energy which will convert entirely into the potential energy of the compressed spring since the surface is frictionless.

The kinetic energy (KE) of the block is calculated using the formula KE = 0.5 × m × v^2, where m is the mass and v is the velocity of the block. The potential energy (PE) stored in the spring at maximum compression is given by PE = 0.5 × k × x^2, where k is the spring constant and x is the compression distance.

Equating the kinetic energy of the block to the potential energy of the spring gives us 0.5 × 3 × (1.5)^2 = 0.5 × 98 × x^2. Solving for x yields x = √((0.5 × 3 × (1.5)^2) / (0.5 × 98)) = 0.229 m or 22.9 cm.

Answer 2

Answer: 0.83 m

Step-by-step explanation:

Given

mass of the block is m=3 kg

spring constant k=98 N/m

The Speed at the time of collision is v=1.5 m/s

Here, the kinetic energy of the block is converted into Elastic potential energy

`\Rightarrow (1)/(2)mv^2=(1)/(2)kx^2 \\\\\Rightarrow 3* 1.5^2=98* x^2\\\\\Rightarrow x^2=0.6887\\\\\Rightarrow x=0.829\approx 0.83\ m`