Physics, 08.11.2019 20:31 mobslayer88

Aboy pulls a sled of mass 5.0 kg with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. the boy pulls on the rope with a force of 10.0 n; and the sled moves with constant velocity. what is the coefficient of friction between the sled and the ice?

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Answer from: drastipatel18

μk = 0.124

Explanation:

Known data

m=5.0 kg : mass of the sled

T= 10 N : force with which the boy pulls the rope

θ =60.0° :angle of the rope with respect to the horizontal direction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law to the sled :

∑F = m*a Formula (1)

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

m : mass s (kg)

a : acceleration (m/s²)

Forces acting on the sled

W: Weight of the sled : In vertical and downward direction

N : Normal force : In vertical and upwards direction

f : Friction force: parallel to the movement of the sled and in the opposite direction to the movement

T:Rope tension : forming angle 60.0° of of the rope with respect to the horizontal direction

Calculated of the W of the sled

W= m*g

W= 5.0 kg* 9.8 m/s² = 49 N

x-y components of the tension of the rope T

Tx= 10*cos60°= 5 N

Ty= 10*sin60° = 8.66 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N+Ty -W = 0

N = 49 N - 8.66 N

N = 40.34 N

Calculated of the f

f = μk* N

f = μk* 40.34 Equation (1)

We apply the formula (1) to calculated f

∑Fx = m*ax the sled moves with constant velocity, then ax=0

∑Fx = 0

Tx-f = 0

5 - f = 0

f = 5N

We replace f in the equation (1)

5 = μk* 40.34

μk = 5 / 40.34

μk = 0.124

Answer from: babygirl091502

The coefficient of friction between the sled and the ice is 0.124.

The given parameters;

mass of the sled, m = 5 kgangle of inclination of the sled, Ф = 60°tension on the rope, T = 10 N

The resultant normal force of the rope is calculated as;

$F_n + Tsin(\theta) - W=0$

$F_n = mg -Tsin(\theta)\\\\F_n = (5\times 9.8) - 10sin(60)\\\\F_n = 40.34 \ N$

The resultant horizontal force on the rope is calculated as;

$Tcos(\theta) -F_k = 0\\\\F_k = Tcos(\theta) \\\\F_k = 10 \times cos (60)\\\\F_k = 5 \ N$

The coefficient of friction between the sled and the ice is calculated as;

$F_k =\mu_k F_n\\\\\mu_k = \frac{F_k}{F_n} \\\\\mu_k = \frac{5}{40.34} \\\\\mu_k = 0.124$

Thus, the coefficient of friction between the sled and the ice is 0.124.

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Answer from: Quest

answer: i don't know the answer good luck

explanation:

Answer from: Quest

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop

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Aboy pulls a sled of mass 5.0 kg with a rope that makes a 60.0° angle with respect to the horizontal...

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