Question

The following two questions (1 and 2) pertain to the diagram shown below.

1- An 0.80kg block is initially compressed against the spring at A by 0.5m. The spring constant of the spring is known to be 25N/m. If the height h illustrated in the diagram is 5.5m, what is the speed of the block when it reaches point B? (The surface between points A and B is frictionless.)
a.) 2.55m/s
b.) 5.80m/s
c.) 8.85m/s
d.) 10.75m/s
e.) 13.35m/s

V1 V4 smooth Baough=C

2- If the coefficient of kinetic friction between points B and E is 0.35, what is the distance from point B to point D (where the block eventually comes to rest), illustrated in the diagram as S?
a.) 6.90m
b.) 9.15m
c.) 11.95m
d.) 14.20m
e.) 16.85m

Answer 1

Answer to part 1:

Once the block is released, it will have a potential energy (due to its position with respect to point B) and Kinetic energy which is imparted to it by the compressed spring (A spring which is compressed by 'x' units have a potential energy equal to $(1/2) *k*x^{2}$ where k is the spring constant).

Therefore,

Total Energy of block at point A = Potential Energy + Kinetic Energy = $(m*g*h) + (1/2)*m*v^{2}$ $\small [Here: (1/2)*m*v^{2} = (1/2)*k*x^{2} = (1/2)*25*(0.5^{2}) = 3.125]$

Total Energy of block at point A = 0.8*10*5.5 + 3.125 = 47.125 J

Now when the block reaches at point B, whole of the Potential energy (m*g*h) at point A is converted to Kinetic Energy at point B.

Total Kinetic energy at point B = Total energy of the block at point A

$\small (1/2)*m*(V_{B})^{2} = 47.125$

$\small (V_{B})^{2} = (47.125*2)/0.8$

$\small (V_{B})^{2} = 117.8$

$\small V_{B} = (117.8)^{1/2}$

$\small V_{B} = 10.85 m/s$

(If g value is taken as 9.8 then answer would be exactly 10.75)

Answer to part 2:

Work Done by friction force from point B to point D = change in kinetic energy from point B to point D

$\small W_{f_{B\rightarrow D}} = K.E_{D} - K.E_{B}$

$\small f_{B\rightarrow D}*Displacement_{B\rightarrow D} = 0 - 47.125$

$\small (\mu*N) *(Displacement_{B\rightarrow D} )= - 47.125$

where u = coefficient of kinetic friction and N is the normal reaction equal to m*g

$\small (0.35*0.8*10) *(Displacement_{B\rightarrow D} )= - 47.125$

$\small (2.8) *(Displacement_{B\rightarrow D} )= - 47.125$

$\small (Displacement_{B\rightarrow D} )= (- 47.125)/2.8$

$\small (Displacement_{B\rightarrow D} )= -16.8 m$

(Negative sign is there because the direction of frictional force and displacement is opposite to each other)

Therefore, Distance from B to D is +16.8 m