Question

A block of mass 500 g is attached to a spring of spring constant 80 N/m. The other end of the spring is attached to a support while the mass rests on a rough surface with a coefficient of friction of 0.20 that is inclined at angle of 30°. The block is pushed along the surface till the spring compresses by 10 cm and is then released from rest. (a) How much potential energy was stored in the block-spring-support system when the block was just released? (b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched. (c) Determine the position of the block where it just comes to rest on its way up the incline.

Answer 1

Answer:

a)  Vki = 0.4 J  

b) v = 0.98 m/s

c)  final position = 0.11745 m from initial position

Explanation:

Given:-

- The mass of block, m = 500 g

- The spring constant, k = 80 N/m

- The coefficient of the surface, u = 0.20

- The inclination of the block, θ = 30°

- The block is initially compressed, xi = 10 cm

- The block is initially at rest, vi = 0 m/s

Find:-

(a) How much potential energy was stored in the block-spring-support system when the block was just released?

(b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched.

(c) Determine the position of the block where it just comes to rest on its way up the incline.

Solution:-

- The potential energy initially stored in a spring "Vki" with constant "k" which has undergone a displacement of "x" is given by:

                            Vki = 0.5*k*xi^2

                            Vki = 0.5*80*0.1^2

                            Vki = 0.4 J  

- The block is released from rest when the energy stored in the spring is dispensed in three forms of energy.

- There would be an increase in the potential energy " Ep " of the block as it moves up.

- There would be an increase in kinetic energy for some time " Ek "

- There would be loss of total energy due to fictitious force ( Ff) the work done against friction will dissipate energy.

- The increase in potential energy "Ep" at displacement xo = 0 from mean position is given by:

                         ΔEp = m*g*Δh ... change in vertical height h

                         ΔEp = m*g*( xi - xo )*sin ( θ )

                         ΔEp = 0.5*9.81*(0.1)*sin ( 30 )

                         ΔEp = 0.24525 J

- The increase in kinetic energy "Ek" at displacement xi = 10 cm, vi = 0 m/s from initial position to mean position at xo = 0 ,its velocity is "vf" is given by:

                         ΔEk = 0.5*m*( vf^2 - vi^2 ) ... change in velocity

                        ΔEk = 0.5*m*( vf )^2

- There would be loss of total energy due to fictitious force ( Ff ) the work done against friction will dissipate energy. First apply the equilibrium conditions on the block normal to slope and determine the contact force ( Nc ):

                        Nc - m*g*cos ( θ ) = 0

                        Nc = m*g*cos ( θ )

- The friction force ( Ff ) is given by:

                        Ff = Nc*u

                        Ff = u*m*g*cos ( θ )

- The work done by block against friction is given by:

                       

                        Wf = -Ff*( xi - xo )

                        Wf = -u*m*g*xi*cos ( θ )

                        Wf = -0.2*0.5*9.81*0.1*cos ( 30 )  

                        Wf = -0.08495 J

                         

- We can now express the work done principle for the block:

                        Vki = ΔEp + ΔEk + Wf

                        ΔEk = - ΔEp - Wf + Vki

                        0.5*m*( vf )^2 = -0.24525 + 0.08495 +0.4

                        vf^2 = 4*(-0.24525 + 0.08495 +0.4 )

                        vf = √0.9588

                        vf = 0.98 m/s

- We will denote the extension of spring at top most position from mean position as "x".

- From mean position xo = 0 m. The block will further move up the slope and expense all its kinetic energy "Ek" in the form of gain in potential energy and gain in elastic potential energy "Vk" and work is done against friction.

               Vki = Ep2 + Wf + Vk2

              0.4 = mg*x*sin ( 30 ) + 0.24525 + 0.08495 + u*m*g*x*cos ( θ ) + 0.5*k*x^2

               0.5*80*x^2 + x*(0.5*9.81*sin(30) + 0.2*0.5*9.81*cos ( 30 ) ) + 0.3302-0.4 =0

               40x^2 + 3.30207x - 0.0698 = 0

Solve the quadratic equation:

               x = -0.1 m (10 cm) - initial compression at rest ;

               x = 0.01745 m

- So the extension in spring at the rest position is x = 0.01745 m. The position of the next resting point is:

                final position = xi + x

                final position = 0.1 + 0.01745

                final position = 0.11745 m from initial position.

               

                x =  

                       

Answer 2

(a) The stored PE will be "0.40 J".

(b) The block's speed will be "0.48 m/s".

(c) The block's position will be at "12 cm".

Potential energy (P.E)

According to the question,

Mass, m = 500 h

Spring constant = 80 N/m

Coefficient of friction = 0.20

Inclined angle = 30°

(a) We know the formula,

Spring P.E = $\frac{1}{2}$ kx²

By substituting the values,

                 = $\frac{1}{2}$ × 80 × (0.1)²

                 = 0.40 J

(b) We know that,

→ $P.E_s$ = ΔKE + Work done by friction + Δ$PE_g$

or,

          = $\frac{1}{2}$ mv² + $\mu_R$ mgx Cosθ + mgx Sinθ

By substituting the values,

 0.40 = $\frac{1}{2}$ × 0.5 × v² + 0.2 × 0.5 × 9.8 × Cos30° × 0.1 + 0.5 × 9.8 × 0.1 ×  Sin30°

          = 0.48 m/s

(c) By using the above formula,

→ 0.40 = $\mu_R$ mgl Cosθ + mgl Sinθ

           = 0.2 × 0.5 × 9.8 × Cos30° × l + 0.5 × 9.8 × l × Sin30°

         l = 0.12 m or,

           = 12 cm

Thus the above responses are correct.

         

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