A metal disk weighing 1 N is resting on an index card that is balanced on top of a glass. When the index card is quickly pulled
2 answers:
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Answer:
a = 0.16
Explanation:
given,
mass of the object 1 = 0.2 kg
mass of the object 2 = 0.3 kg
acceleration when force is on 0.2 kg = 0.4 m/s²
acceleration when both mass are combine = ?
F = m a
F = 0.2 × 0.4
F = 0.08 N
force acting is same and total mass = 0.2 + 0.3 = 0.5 Kg
F = m a
a = 0.16 m/s²
the acceleration acting when both the body is attached is a = 0.16
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m