Question

A boy 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?

Answer 1

Answer:

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