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2 years ago

How much kinetic energy does a person of mass 50kg running at 3m/s have? Provide

Physics

1 answer:

ddd [48]2 years ago

4 0

Answer:

0.225 KJ kinetic energy is used for a person of mass 50 Kg running at speed of 3 m/s.

Explanation:

Given:

mass = 50 kg

velocity = v = 3 m/s

To Find :

Kinetic energy = ?

Solution:

Kinetic energy of an object is the energy that it possesses due to its motion.

The formula for kinetic energy is given as

$\textrm{kinetic energy} = \frac{1}{2}\times mass\times velocity^{2}\\= \frac{1}{2}\times m\times v^{2}\\= \frac{1}{2}\times 50\times 3^{2}\\= 25\times 9\\= 225\ Joules.$

But we required kinetic energy in KiloJoules i.e KJ

1 K = 1000 = $10^{3}$

$\textrm{kinetic energy} = 225\ J\\= 0.225\times 1000\\= 0.225 \times 10^{3}\\= 0.225\ KJ$

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A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

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Average friction force,f= $4.0\times 10^3 N$

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Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

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2 years ago

A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the

marin [14]

Answer:

0.395 m

2.4 m/s

Explanation:

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$x$ = initial position of spring from equilibrium position = 0.21 m

$v_{i}$ = initial speed of the cart = 2.0 ms⁻¹

$A$ = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

$(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m$

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$A$ = amplitude of the oscillation = 0.395 m

$v_{o}$ = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

$(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}$

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