Question

A boy pushes his little sister in a sled across the horizontal ground. He pushes down and to the right on the sled at an angle of 35 degrees from the horizontal. The mass of the sled and the sister together is 60 kg. The coefficient of static friction of the sled with the ground is 0.3 and the coefficient of kinetic friction is 0.2. a) Draw a sketch of the situation and a force diagram of the sled, labeling all forces clearly. b) With what force does the boy have to push to start the sled moving from rest? c) If he continues to push with this force after the sled is moving, what will be the sled's acceleration? d) How long will he have to push the sled with this force unitil the it is moving at 2 m/s? e) Discuss whether or not your answers to parts b through d pass the "common sense" test. Be specific about why you think the answers do or do not make sense.

Answer 1

given that :

mass of the sled and the sister together, M = 60 kg

coefficient of static friction of the sled with the ground, _s = 0.3

coefficient of kinetic friction, _k = 0.2

making an angle from the horizontal surface, = 35 degree

(a) To draw a free body diagram of the sled moving from rest which can be shown in figure below ::

(b) the boy have to push to start the sled moving from rest by a frictional force which is given as::

from a figure,

f_s = _s F_N                { eq. 1 }

where, F_N = mg

let x be the horizontal axis & y be the vertical respectively.

force components on x-axis, F_x = F Cos                        { eq. 2 }

force components on y-axis, F_y = F Sin                     { eq. 3 }

on vertical plane,

F_y = F_N + F Sin - mg            { eq.4 }

inserting the values in eq.4,

0 = F_N + F Sin (35) - (60 kg) (9.8 m/s²)   

F_N = - (0.574) F + 588 N                               { eq.5 }

on horizontal plane :

F_x = F Cos - f_s                  { eq. 6 }

inserting the values in eq.6,

F_x = F Cos (35) - _s F_N    

0 = F Cos (35) - (0.3) [- (0.574) F + 588 N]

0 = (0.819) F - (0.017) F - 176.4 N

176.4 = (0.802) F

F = 220 N

(c) If he continues to push with this force after the sled is moving, then sled acceleration is given as ::

ma_x = F - f_s

or   a_x = F - f_s / m                         { eq.7 }

inserting the values in eq.7

a_x = [(220 N) - (0.017 x 220 - 176.4)] / (60 kg)

a_x = (392.6) / (60)

a_x = 6.54 m/s²

(d) he have to push the sled with this force unitil the it is moving at 2 m/s, then time taken is given as :

using equation of motion 1,

v = u + at                    { eq.8 }

where, v = 2 m/s

inserting the values in above eq.

(2 m/s) = 0 + (6.54 m/s²) t

t = (2 m/s) / (6.54 m/s²)

t = 0.3 sec