Question

Problem 5 A block of mass 3 kg slides on a horizontal, rough surface towards a spring with k = 500 N/m. The kinetic friction coefficient between the block and the surface is µk = 0.6. If the block’s speed is 5 m/s at the instant it first makes contact with the spring, (a) Find the maximum compression of the spring. (b) Draw work-energy bar diagrams for the process of the block coming to a halt, taking the system to be the block and the surface only

Answer 1

Answer:

Explanation:

given,

mass of block = 3 kg

spring constant k = 500 N/m

kinetic friction coefficient µk = 0.6

speed of block = 5 m/s

F = µk  N

F = 0.6 x 3 x 9.8

F = 17.64 N

using energy conservation

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2+Fx$

$\dfrac{1}{2}\times 3 \times 5^2=\dfrac{1}{2}\times 500 \times x^2+17.64\times x$

250 x² + 17.64 x - 37.5 = 0

on solving

x = 0.354 m

graph is attached below

Answer 2

Explanation:

Given that,

Mass of block = 3 kg

Spring constant k=500 N/m

Friction coefficient = 0.6

Speed = 5 m/s

(a). We need to calculate the maximum compression of the spring

Using work energy theorem

$\dfrac{1}{2}mv^2+\mu\times mgx=\dfrac{1}{2}kx^2$

Put the value into the formula

$\dfrac{1}{2}\times3\times(5)^2+0.6\times3\times9.8\times x=\dfrac{1}{2}\times500\times x^2$

$250x^2-17.64x-37.5=0$

$x=-0.354\ m$

Negative sign shows the compression.

The maximum compression of the spring is 0.354 m.

(b). We need to draw the bar diagram

We need to calculate the initial energy

$E_{i}=\dfrac{1}{2}kx^2$

Put the value into the formula

$E_{i}=\dfrac{1}{2}\times500\times(0.354)^2$

$E_{i}=31.329\ J$

We need to calculate the final energy

Using formula of final energy

$E_{f}=\dfrac{1}{2}mv^2$

$E_{f}=\dfrac{1}{2}\times3\times(5)^2$

$E_{f}=37.5\ J$

We need to calculate the work

Using formula of work

$W=Fx$

$W=\mu mg\times x$

Put the value into the formula

$W=0.6\times3\times9.8\times(-0.354)$

$W=-6.244\ J$

Hence, This is the required solution.