Question

A 1.50-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?

Answer 1

0.032 m

Step-by-step explanation:

Consider the forces acting on the block

m = mass of the block = 1.50 kg

`f_(s)` = Static frictional force

`F_(n)` = Normal force on the block from the wall

`F_(s)` = Spring force due to compression of spring

`F_(g)` = 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

`F_(s)` = kx

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

`F_(n)` =
`F_(s)`

`F_(n)` = kx eq-1

Static frictional force is given as

`f_(s)` = μ
`F_(n)`

Using eq-1

`f_(s)` = 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

`f_(s)` =
`F_(g)`

Using eq-2

μ k x = 14.7

(0.54) (860) x = 14.7

x = 0.032 m