Answer:
a) 0.11825
b) 3.401 m/s
c) 10.72 m
d) 2.47 m
Explanation:
Given:-
- The mass of the block, m = 3.7 kg
- The angle of the inclined plane, θ = 51°
- The spring constant of inclined spring, k1 = 31 N/m
- The spring constant of horizontal spring, k2 = 6.9 N/m
- The coefficient of friction ( inclined plane ) μ1 = 0.24
- The coefficient of friction ( horizontal surface ) μ2 = 0.055
- The initial compression of inclined spring, Δx = 0.11 m
- The initial height of the block, h = 0.64 m
- The horizontal distance to second spring, d = 0.14 m
a)
Solution:-
- We will denote the distance along the inclined surface as "Δs" that the block travels when released from its initial position: ( xi = Δx ).
- The final position where the block on the inclined plane comes to a stop is x = x1. The displacement Δs can be written as:
Δs = xi - x
Δs = Δx - x1
Note: The equilibrium position is considered as the origin.
The work-done by the block against friction ( W ):
W = F*Δs
W = u*m*g*cos ( θ )*Δs
- Use a energy balance for the block between the initial compressed point and the final point on the inclined surface where the block comes to a stop:
U1 + Ep1 = Ep2 + W
Where,
U1 : The elastic potential energy = 0.5*k1*Δx^2
Ep1: The initial gravitational potential energy = m*g*( h + Δx*sin ( θ ) )
Ep2: The final gravitational potential energy = m*g*( h1 )
- The change in gravitational potential energy ΔEp = Ep2 - Ep1:
ΔEp = Ep2 - Ep1 = -m*g*( h + Δx*sin ( θ ) ) + m*g*( h1 )
= m*g*( (h1 - h) - Δx*sin ( θ ) )
= m*g*( -x1*sin ( θ ) - Δx*sin ( θ ) )
= m*g*sin ( θ )* ( -x1 - Δx )
= - m*g*sin ( θ )*( Δs )
- Use the energy principle expression stated above and solve for Δs:
0.5*k1*Δx^2 = - m*g*sin ( θ )*( Δs ) + u1*m*g*cos ( θ )*Δs
0.5*k1*Δx^2 = Δs [ m*g* ( u1*cos ( θ ) - sin ( θ ) ) ]
0.5*31*0.11^2 = Δs [ 3.7*9.81* ( 0.24*cos ( 51° ) - sin ( 51° ) ) ]
0.18755 = -22.72588*Δs
Δs = - 0.00825 m
The x-coordinate of the resting point would be:
Δs = Δx - x1
x1 = Δx - Δs
x1 = 0.11 - ( -0.00825 )
x1 = 0.11825 m
b)
Solution:-
Use a energy balance for the block between the initial compressed point and the final point at the bottom of inclined surface where the block has a velocity "u":
U1 + Ep1 = W + Ek + Ep2
Where,
U1 : The elastic potential energy = 0.5*k1*Δx^2
Ep1: The initial gravitational potential energy = m*g*( h + Δx*sin ( θ ) )
Ek: The kinetic energy at the bottom of inclined surface = 0.5*m*u^2
Note: Taking the horizontal surface as the datum ( Ep2 = 0 )
Therefore,
0.5*k1*Δx^2 + m*g*( h + Δx*sin ( θ ) ) = u1*m*g*cos ( θ )*Δs + 0.5*m*u^2
Where,
Δs: The total distance from initial point to bottom surface = Δx + h / sin ( θ )
0.5*k1*Δx^2 + m*g*( h + Δx*sin ( θ ) ) = u1*m*g*cos ( θ )*(Δx + h / sin ( θ )) + 0.5*m*u^2
0.5*31*0.11^2 + 3.7*9.81*( 0.64 + 0.11*sin ( 51° ) ) = 0.24*3.7*9.81*cos ( 51° )* ( 0.11 + 0.64 / sin ( 51° ) ) + 0.5*3.7*u^2
0.18755 + 26.33296 = 5.11776 + 1.85*u^2
21.40275 = 1.85*u^2
u = √( 21.40275 / 1.85 ) = √11.56905
u = 3.401 m/s .... Answer
c)
Solution:-
- Use a energy balance for the block between the point at the bottom of the inclined surface and the final point where block reaches its maximum distance "s" by doing work against friction:
Ek = W1
0.5*m*u^2 = μ2*m*g*s
0.5*u^2 = μ2*g*s
s = 0.5*3.401^2 / ( 0.055*9.81)
s = 10.71893 m ≈ 10.72 m ... Answer
d)
Solution:-
- The new friction force acting on the block acts for the distance of d = 0.14m.
- The initial kinetic energy of the block corresponding to the speed ( u ) at the bottom of the inclined surface is reduced to speed ( v ) due to loss of kinetic energy by working against the friction.
- Apply the work-done principle against the new friction over the distance d travelled by the block on the horizontal surface is expressed as:
Ek1 = W2 + Ek2
0.5*m*u^2 = μ2*m*g*d + 0.5*m*v^2
- The final velocity of block ( v ) after doing work against the friction ( W2 ) can be determined by the above expressed energy principle:
u^2 - 2*μ2*g*d = v^2
v^2 = 3.40133^2 - 2*(0.055)*(9.81)*(0.14)
v = √11.41797
v = 3.37904 m/s
- After doing work against the friction the block's kinetic energy is stored into the spring in the form of elastic potential energy ( U2 ).
- The conservation of energy principle can be applied ( No fictitious work done ).
Ek = U2
0.5*m*v^2 = 0.5*k2*Δs^2
Δs = v*√(m / k2) = 3.37904*√(3.7 /6.9) =
Δs = 2.47440 m ≈ 2.47 m ... Answer