Boy pulls a 5.0-kg sled 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 and the ice?

(a) 0.09
(b) 0.12
(c) 0.18
(d) 0.06
(e) 0.24

0.1

Explanation:

mass, m = 5 kg

θ = 60°

Force, F = 10 N

velocity is constant , it means the net force is zero.

So, the component of force along the surface is equal to the friction force

FCosθ = friction force

10 x cos 60 = μ x m x g

where, μ is the coefficient of friction

5 = μ x 5 x 9.8

μ = 0.1

Thus, the coefficient of friction is 0.1

wvbbvvv

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s