Question

A 40 kg block is pushed up against a spring as shown. The spring constant is 400N/m and the coefficient of static friction is .25. How far must the spring be compressed in order to have enough force to overcome static friction and push the block into motion?

Answer 1

Force due to friction=Force due to spring

u*m*g=K*x

0.25*40*9.81=400*x

x=0.2452 m

=24.52 cm ans

Answer 2

The maximum force of static friction that can be applied by ground

$= \mu m g = 0.25 \times 40 \times 9.8 N = 98 N$

To balance that let required compression in spring = x, hence force by spring is kx.

$\Rightarrow 400x = 98$

$\Rightarrow x = \frac{98}{400} = \frac{49}{200} metres$   

Answer 3

Alagau si indicats com paess ion n is inm Assuming bakk is on hni3cmtzd plane 0.481 (0+ 025 -0.24 525m