An adventurer pulls a loaded sled with a mass of 56kg along a horizontal meadow at a constant speed. the coefficient of kinetic friction between the sled and the grass is 0.45 and the angle of the rope between the sled and the adventurer is 55 degrees. what is the tension in the rope?

he tension is 262 N

Explanation:

1) Call T the tension of the rope

2) Split T into its horizontal and vertical components.

3) Horizontal component of the tension, Tx = T cos(55°)

4) Vertical component of the tension, Ty = T sin (55°)

5) Force equilibrium in the vertical direcction:

∑Fy = 0

Ty + normal force - weight of the sled = 0

Call N the normal force

Ty + N - 56 kg * 9.8 m/s^2

Ty + N = 56 kg * 9.8 m/s^2

Ty + N = 548.8N

T sin(55) + N = 548.8N

6) Force equilibrium in the horizontal direction

constant velocity => ∑Fx = 0

Tx - Fx = 0

Tcos(55) - Fx = 0

7) Fx is the friction force.

The friction force and the normal force are related by the kinetic friction coefficient.

Call μk the friction coefficient

Fx = μk N

=> Tcos(55) - μk N = 0

Tcos(55) - 0.45N = 0

8)  Solve the system of two equations:

Eq (1) T sin(55) + N = 548.8
Eq (2) T cos(55) - 0.45N = 0

Eq(1) 0.819T + N = 548.8
Eq(2) 0.574T - 0.45N = 0

The solution of the system is T = 262.01 N and N = 334.21 N

Then T ≈ 262N

Ithink the answer would be 1,350!

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward