Question

You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.4 m away, asks you for the box of nails. Rather than carry the 2.1 kg box of nails down to Peter, you decide to give the box a push and have it slide down to him. If the coefficient of kinetic friction between the box and the roof is 0.51, with what speed should you push the box to have it gently come to rest right at the edge of the roof?

Answer 1

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : 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 weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box 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  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

$v_{o} =\sqrt{14.472}$

v₀ = 3.8 m/s