Question

A 2.00 kg textbook is forced against one end of a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a distance of 0.300 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction μk = 0.30.

Answer 1

Answer:

d = 1.875 m

Explanation:

Given that

m = 2 kg

k = 250 N/m

x= 0.3 m

Lets take the speed of book just before when it is leaving the spring

From energy conservation

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

$kx^2=mv^2$

250 x 0.3²= 2 x v²

v= 3.35

When it moves on the rough surface then de -acceleration a

a=  μk g

a= 0.3 x 10 = 3 m/s²

The distance cover by book before getting to rest position is d

The final speed(V) of the block will be zero.

We know that

V² = v²- 2 a d

0 = 125 x 0.3²  - 2 x 3 x d

d = 1.875 m