A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ

4 answers
Question:

A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position. Required:
a. Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive.
b. Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression ΣFx = 25%
c. Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1 m?
d. Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?

Answers

Explanation:

given

spring constant k = 880 N/m

mass m = 150 kg

Normal force will be equal to the component of weight of mass m which is perpendicular to the inclined surface

= mgcosθ

So normal force

FN = mgcosθ j , as it acts in out of plane direction .

b )

Fricrion force acting in upward direction = μs mgcosθ

component of weight acting in downward direction = mgsinθ

restoring force by spring on block in downward direction

= kd

= 880d

F( total ) = (μs mgcosθ -  mgsinθ -  880d )i

c )  

for balance

mgsinθ =  kd

sinθ = kd / mg

= 880 x .1 / 150 x 9.8

= 88 / 1470

.0598

θ = 3.4 degree

d )

d = mgsinθ / k

150 x 9.8 sin45 / 880

= 1.18 m

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

             

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = [tex]\frac{k \Delta x}{ m g}[/tex]

            cos θ = [tex]\frac{880 \ 0.1}{ 150 \ 9.8}[/tex]

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = [tex]\frac{mg \ cos \ \theta}{k}[/tex]

            Δx = [tex]\frac{ 150 \ 9.8 \ cos45}{880}[/tex]

            Δx = 1.18 m


[tex]A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclin[/tex]

not too sure thinking all accelerations are pointed down

Scissors and a seesaw both cut things that’s the only comparison

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