Respuesta :
Answer: 0.374 m/s
Explanation: The motion of a loaded spring is assumed to be of a constant acceleration, hence newton's laws of motion is applicable.
Recall from the second law of motion that
F - Fr = ma
Where F = applied force = elastic force = kx
Fr = frictional force =?
u = coefficient of kinetic friction = 0.39
m = mass of block = 2.60kg
g = acceleration due to gravity = 9.8 m/s²
k = elastic constant = 810 N/m
x = 0.0310m
Fr = umg
Fr = 0.39 × 2.60 × 9.8
Fr = 9.94.
F = kx = 810 × 0.0310m = 25.11 N
By substituting the parameters, we have that
25.11 - 9.94 = 2.60 (a)
15.17 = 2.60a
a = 15.17/ 2.60 = 5.84 m/s².
Speed of block (v) =?
Distance covered (s) = 0.0120m
Initial velocity (u) = 0
Recall that
v² = u² + 2as
v² = 2as
v² = 2 × 5.84 × 0.0120m
v² = 0.1400
v = √0.1400
v = 0.374 m/s
Answer:
0.4323 m/s
Explanation:
We are given;
Mass; m = 2.60kg
Spring Constant; k = 810N/m
coefficient of kinetic friction; μ_k = 0.39
When we apply Newton's second law of motion to the block in the vertical direction, we get;
Σy = N - mg = 0
So, N = mg
Mow,we know that formula for kinetic friction is;
F_k = μ_k•mg
Thus, F_k = 0.39 x 2.6 x 9.8 = 9.9372N
Now, we know that; work = Force x distance.
Since the block when it has moved a distance of 0.0120m from its initial position. Work done by frictional force is;
W_f = - 9.9372 x 0.012 = - 0.119J
The negative sign implies that the frictional force is opposite to the direction of displacement.
Formula for kinetic energy is; K = (1/2)mv²
While elastic potential Energy is given as;
U = (1/2)kx²
Thus,at initial point,
K1 = (1/2) x 2.6 x 0² = 0
U1 = (1/2) x 810 x 0.031² = 0.3892 J
At second point;
K2 = (1/2) x 2.6 x (v_2)² = 1.3(v_2)² J
U2 = (1/2) x 810 x 0.019² = 0.1462J
Work energy theorem gives that;
K1 + U1 = K2 + U2
Thus,
0 + 0.3892 = 1.3(v_2)² + 0.1462
1.3(v_2)² = 0.3892 - 0.1462
(v_2)² = 0.243/1.3
v_2 =√0.1869 = 0.4323 m/s
