Answer:
21 m
Explanation:
Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.
Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N
The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N
The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N
If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.
Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.
For the block to be in contact with the surface, the vertical forces on the block must balance.
Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,
N = mg + F₃" (since both the weight and the resultant vertical force act downwards)
N = mg + F₃"
Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N
So,
N = mg + F₃"
N = 35 kg × 9.8 m/s² + 8.82 N
N = 343 N + 8.82 N
N = 351.82 N
So, the net horizontal force F = F₃' - f.
F = 155.41 N - 0.4 × 351.82 N
F = 155.41 N - 140.728 N
F = 14.682 N
Since F = ma, where a = acceleration of block,
a = F/m = 14.682 N/35 kg = 0.42 m/s²
To find the distance the block moved, x we use the equation
x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²
Substituting the values of the variables into the equation, we have
x = ut + 1/2at²
x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²
x = 0 m + 1/2 × 0.42 m/s² × 100 s²
x = 0.21 m/s² × 100 s²
x = 21 m
So, the distance moved by the block is 21 m.
