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9 days ago

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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 w

hen 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?

1 answer:

Keith_Richards [1K]9 days ago

7 0

Answer:

v₀ = 3.8 m/s

Explanation:

According to Newton's second law relating to the box:

∑F = m*a Formula (1)

∑F: the net force in Newton (N)

m: mass expressed in kilograms (kg)

a: acceleration measured in meters per second squared (m/s²)

Information known:

m = 2.1 kg, the mass of the box

d = 5.4m, the length of the roof

θ = 20° is the angle between the roof and the horizontal

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

g = 9.8 m/s², gravitational acceleration

Forces influencing the box:

The x-axis is oriented parallel to the box's movement on the roof, and the y-axis is oriented perpendicularly.

W: Weight of the box: directed vertically

N: Normal force: perpendicular to the roof's angle

fk: Frictional force: parallel to the direction along the roof

Calculating the weight of the box:

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

x-y components of weight:

Wx= Wsin θ=(20.58)*sin(20)°=7.039 N

Wy= Wcos θ=(20.58)*cos(20)°= 19.34 N

Finding the Normal force:

∑Fy = m*ay ay = 0

N-Wy = 0

N=Wy = 19.34 N

Calculating the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We substitute into Formula (1) to determine the box's acceleration:

∑Fx = m*ax ax=a: acceleration of the box

Wx-fk = (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

Considering the box's Kinematics:

Since the box undergoes uniformly accelerated motion, we use the following to find the final speed of the box:

vf² = v₀² + 2*a*d Formula (2)

Where:

d refers to displacement = 5.4 m

v₀ is the initial speed

vf represents the final speed = 0

a is the box's acceleration = -1.34 m/s²

Plugging in the values into 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

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