A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
1 answer:
The box has 3 forces acting on it:
• its own weight (magnitude <em>w</em>, pointing downward)
• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)
• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>
• net perpendicular force:
∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
<em>n</em> = <em>w</em> cos(35°)
<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)
<em>n</em> ≈ 120 N
Solve for the mag. of friction:
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.25 (120 N)
<em>f</em> ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>
<em>a</em> ≈ (54.3157 N) / (15 kg)
<em>a</em> ≈ 3.6 m/s²
Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:
<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>
<em>v</em>² = 2 (3.6 m/s²) (3 m)
<em>v</em> = √(21.7263 m²/s²)
<em>v</em> ≈ 4.7 m/s
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Answer:
a) F₁ = 1.48 x 10³ N
b) P = 1.88*10⁵ Pa
c) The work is equal in both pistons
Explanation:
Pascal´s Principle can be applied in the hydraulic press:
If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.
Pressure is defined as the force per unit area:
Formula (1)
P₁=P₂
Formula (2)
Data
r₁= 5 cm = 0.05 m
r₂= 15 cm = 0.15 m
F₂= 13300N
Area of the pistons (A₁,A₂)
A=π*r² : Area of the circle
A₁ = π*(0.05)²=7.85*10⁻³ m²
A₂= π*(0.15)²= 70.69*10⁻³ m²
a) Force that compressed air must exert to lift a car weighing 13300 N
We replace data in the formula (2)
F₁ = 1.48 x 10³ N
b) Air pressure produced by F₁
We replace data in the formula (1)
F₁ = 1.48 x 10³ N , A₁ = 7.85*10⁻³ m²
P= 1.88*10⁵ Pa
c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.
Answer:
total kinetic energy is 8 × J
Explanation:
given data
potential difference = 5 V
e = 1.60 × C
to find out
what is kinetic energy
solution
we will apply here conservation of energy that is
change in potential energy is equal to change in kinetic energy
so
change potential energy is e × potential difference
change potential energy = 1.60 × × 5
change potential energy = 8 × J
so change in kinetic energy = 8 × J
and we know proton start from rest that mean ( kinetic energy is 0 ) so
change in KE is total KE
total kinetic energy is 8 × J