A1.50-kg block is pushed against a vertical wall by means of a spring (k = 860 n/m). the coefficient of static friction between the block and the wall is 0.54. what is the minimum compression in the spring to prevent the block from slipping down?

0.032 m

Explanation:

Consider the forces acting on the block

m = mass of the block = 1.50 kg

= Static frictional force

= Normal force on the block from the wall

= Spring force due to compression of spring

= Force of gravity on the block = mg = 1.50 x 9.8 = 14.7 N

k = spring constant = 860 N/m

μ = Coefficient of static friction between the block and wall = 0.54

x = compression of the spring

Spring force is given as

= kx

From the force diagram of the block, Using equilibrium of force along the horizontal direction, we get the force equation as  

=

= kx                                             eq-1

Static frictional force is given as

= μ

Using eq-1

= μ k x                                                eq-2

From the force diagram of the block, Using equilibrium of force along the vertical direction, we get the force equation as

=

Using eq-2

μ k x = 14.7

(0.54) (860) x = 14.7

x = 0.032 m

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures