405 kg penguin sits on a 100 kg sled as seen in the figure b

4.05 kg penguin sits on a 10.0 kg sled, as seen in the figure below the motion by holding onto a cord attached to a tree. The coeffmicient of kinetic friction between the sled and A horizontal force of F- 56.9 N Is applied to the sied, but the penguin attempts to impede snow as well as that between the sled and the penguin is 0.175. Determine the tension in the cord in N) A 37:434C: 5.07D: 5.94E: 6.95 OF8.13G 9.51N: 1.1110 Sumit Anwr Tries 0/3 Determine the acceleration of the sled (in m/s 2) A: 4.67x10 B: 6.21x10C: 8.26×10-11 1.1 E: 1.46|-F: 1.941 G: 2.59:H: 3.44

Solution

A] We know that,

F = ma

where,F = the sum of all forces acting on the object

m = mass of the object

a = acceleration of the object.

The tension in the cord and the force of friction (Ffr1) between the sled and the penguin, are the only two forces acting along the x-axis. So by modifying the above equation:

Fx = mp(ap)

where, Fx = sum of all forces acting on the penguin along the x-axis = -Tension + Ffr1

mp = mass of the penguin = 4.9 kg

ap = acceleration of penguin = 0

Ffr1 = (coefficient of friction)(Normal Force)

Normal Force = Weight of the penguin

Weight of the penguin = mp(g)

g = gravity = 9.8 m/s^2

You find that the tension equals the force of friction in the opposite direction:

-T = -(0.175)*(4.05kg)*(-9.8 m/s^2)

T = - 6.945 N

B] By using the equation

Fx = mt(as)

where, Fx = sum of all forces acting on the sled along the x-axis

mt = total mass of the sled and the penguin = 10+4.05 = 14.05 kg

as = acceleration of the sled = unknown

A sum of all the forces gives you

F + Ffr1 + Ffr2 = mt(as)

where, F = applied force = 56.9 N

Ffr1 = force of friction between penguin and sled = - 6.945 N

Ffr2 = force of friction between sled and snow

Ffr2 = (coeff of friction)(Normal Force)

Normal Force = mt(g) = 14.05*9.8 = - 137.69 N

So,

56.9 N - 6.945 N – (137.69 N)*(0.175) = 14.05 kg * (as)

as = {56.9 N - 6.945 N – (137.69 N)*(0.175) }/14.05

= 1.84 m/s^2 answer