Assuming the woman starts at rest, she descends the slide with acceleration <em>a</em> such that
(20.3 m/s)² = 2 <em>a</em> (42.6 m) → <em>a</em> ≈ 4.84 m/s²
which points parallel to the slide.
The only forces acting on her, parallel to the slide, are
• the parallel component of her weight, <em>w</em> (//)
• friction, <em>f</em>, opposing her descent and pointing up the slide
Take the downward sliding direction to be positive. By Newton's second law, the net force in the parallel direction acting on the woman is
∑ <em>F</em> (//) = <em>w </em>(//) - <em>f</em> = <em>ma</em>
where <em>m</em> = 77.0 kg is the woman's mass.
Solve for <em>f</em> :
<em>mg</em> sin(42.3°) - <em>f</em> = <em>ma</em>
<em>f</em> = <em>m</em> (<em>g</em> sin(42.3°) - <em>a</em>)
<em>f</em> = (77.0 kg) ((9.80 m/s²) sin(42.3°) - 4.84 m/s²) ≈ 135 N
Compute the work <em>W</em> done by friction: multiply the magnitude of the friction by the length of the slide.
<em>W</em> = (135 N) (42.6 m) ≈ 5770 N•m = 5770 J
Answer:
9.6 m
Explanation:
This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .
a = 3.6 t + 5.6
d²x / dt² = 3.6 t + 5.6
Integrating on both sides
dx /dt = 3.6 t² / 2 + 5.6 t + c
where c is a constant.
dx /dt = 1.8 t² + 5.6 t + c
when t = 0 , velocity dx /dt is zero
Putting these values in the equation above
0 = 0 +0 + c
c = 0
dx /dt = 1.8 t² + 5.6 t
Again integrating on both sides
x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁
x = 0.6 t³ + 2.8 t² + c₁
when t =0, x = 0
c₁ = 0
x = 0.6 t³ + 2.8 t²
when t = 1.6
x = .6 x 1.6³ + 2.8 x 1.6²
= 2.4576 + 7.168
= 9.6256
9.6 m
Explanation:
A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,
h = 2.3 m
So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
kinetic <span>energy when the pitcher has thrown it or when one of the other players has thrown the ball. The baseball also has </span>kinetic<span> energy when the batter hits the ball. when you catch the ball it is potential energy.</span>
