Answer:
D
Explanation:
I hope you get a good grade!
C,d,or e you can use the process of elimination to decide...
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer: D. ➡️⬅️
Explanation: I just knew the answer ;)
Answer:
Option E is correct 310N
Explanation:
Given that the force used to push the crate is F = 200N
The force directed 20° below the horizontal
Mass of crate is m = 25kg
Weight of the crate can be determine using
W = mg
g is gravitational constant =9.8m/s²
W = 25×9.8
W = 245 N
Check attachment. For free body diagram and better understanding
Using newton second law along the vertical axis since we want to find the normal force
ΣFy = m•ay
ay = 0, since the body is not moving in the vertical or y direction
N—W—F•Sin20 = 0
N = W+F•Sin20
N = 245+ 200Sin20
N = 245 + 68.4
N = 313.4 N
The normal force is approximately 310 N to the nearest ten
