Answer:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N - =
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
Work done by the car is 3900 J
Explanation:
- Power and work are related by the equation, Power = Work Done/Time
- Power is the rate at which work is done.
- Here, the car uses power of 260 W and time taken is 15 s.
Work Done = Power × Time
= 260 × 15 = 3900 J
Its b.functional paragraph because writers use this for interest presents and special effects
Answer:
μ = 0.18
Explanation:
Let's use Newton's second Law, the coordinate system is horizontal and vertical
Before starting to move the box
Y axis
N-W = 0
N = W = mg
X axis
F -fr = 0
F = fr
The friction force has the formula
fr = μ N
fr = μ m g
At the limit point just before starting the movement
F = μ m g
μ = F / m g
calculate
μ = 34.8 / (19.8 9.8)
μ = 0.18