While a roofer is working on a roof that slants at θ = 36.0 degrees above the horizontal, he accidentally nudges his m = 8.50 kg toolbox, causing it to start sliding downward, starting from rest. a frictional force of magnitude fk = 22.0 n acts on the toolbox as it slides. if the box starts d = 4.25 m from the lower edge of the roof, how fast v will the toolbox be moving just as it reaches the edge of the roof? assume that the acceleration due to gravity is g = 9.80 m/s2

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

vf= 5.19 m/s

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

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