Answer:
d = 11.79 m
Explanation:
Known data
m=4.25 kg : mass of the block
θ =37.5° :angle θ of the ramp with respect to the horizontal direction
μk= 0.460 : coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.
Forces acting on the block
W: Weight of the block : In vertical direction
N : Normal force : perpendicular to the ramp
f : Friction force: parallel to the ramp
Calculated of the W
W= m*g
W= 4.25 kg* 9.8 m/s² = 41.65 N
x-y weight components
Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N
Wy= Wcos θ =41.65*cos 37.5° =33.04 N
Calculated of the N
We apply the formula (1)
∑Fy = m*ay ay = 0
N - Wy = 0
N = Wy
N = 33.04 N
Calculated of the f
f = μk* N= 0.460*33.04
f = 15.2 N
We apply the formula (1) to calculated acceleration of the block:
∑Fx = m*ax , ax= a : acceleration of the block
-Wx-f = m*a
-25.35-15.2 = (4.25)*a
-40.55 = (4.25)*a
a = (-40.55)/ (4.25)
a = -9.54 m/s²
Kinematics of the block
Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement (m)
v₀: initial speed (m/s)
vf: final speed (m/s)
Data:
v₀ = 15 m/s
vf = 0
a = -9.54 m/s²
We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d)
vf²=v₀²+2*a*d
0 = (15)²+2*(-9.54)*d
2*(9.54)*d = (15)²
(19.08)*d = 225
d = 225 / (19.08)
d = 11.79 m
