Answer:
vf= 5.19 m/s
Explanation:
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/s
Known data
m = 4.25 kg : mass of the toolbox
θ =36.0° : angle θ of the roof with respect to the horizontal direction
fk = 22.0 N : magnitude of the frictional force
g = 9.8 m/s² : acceleration due to gravity
Forces acting on the block
We define the x-axis in the direction parallel to the movement of the toolbox on the roof and the y-axis in the direction perpendicular to it.
W: Weight of thetoolbox : In vertical direction
N : Normal force : perpendicular to the direction the roof
fk : Friction force: parallel to the direction to the roof
Calculated of the W
W = m*g
W = 8.50 kg* 9.8 m/s² = 83.3 N
x-y weight components
Wx= Wsin θ= 83.3 N*sin(36)° = 48.96 N
Wy= Wcos θ =83.3 N*cos(36)° =67.39 N
Acceleration of the toolbox
We apply formula (1) to the toolbox to calculate its acceleration :
∑Fx = m*ax , ax= a : acceleration of the toolbox
Wx-fk = m*a
48.96 -22 = (8.5)*a
26.96 = (8.5)*a
a = (26.96 ) / (8.5)
a = 3.17 m/s²
Kinematics of the toolbox
Because the toolbox moves with uniformly accelerated movement we apply the following formula to calculate its the final speed :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement (m)
v₀: initial speed (m/s)
vf: final speed (m/s)
Data:
v₀:0
d = 4.25 m
a = 3.17 m/s²
We replace the data in formula (2) to calculate the speed of the toolbox just when it reaches the edge of the roof :
vf²=v₀²+2*a*d
vf²=0+(2)*(3.17)* (4.25)
vf²=26.945
[tex]v_{f} =( \sqrt{26.945} )\frac{m}{s}[/tex]
vf= 5.19 m/s
