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belka [17]
3 years ago
15

A 2.36 kg block resting on a frictionless surface is attached to an ideal spring with spring constant k = 260 Nm . A force is ap

plied to the block which changes its position from xi=5.89 cm to xf=−15.4 cm , each distance measured relative to the equilibrium position of the block. While the block is being moved, find
(a) the work done by the spring and
(b) the work done by the applied force.
Physics
1 answer:
podryga [215]3 years ago
8 0

Answer:

-2.63 Joules

2.63 Joules

Explanation:

x_i = Initial compression = 5.89 cm

x_f = Final compression = -15.4 cm

k = Spring constant = 260 Nm

Work done by a spring is given by

W=\frac{1}{2}k(x_i^2-x_f^2)\\\Rightarrow W=\frac{1}{2}260\times (0.0589^2-0.154^2)\\\Rightarrow W=-2.63\ J

Work done by the spring is -2.63 Joules.

Change in kinetic energy is given by

\Delta K=W_a+W_s

Here, it is assumed that change in kinetic energy is zero as velocity and amlitude are not mentioned.

So,

0=W_a+W_s\\\Rightarrow W_a=-W_s\\\Rightarrow W_a=--2.63\\\Rightarrow W_a=2.63\ J

The work done by the applied force is 2.63 Joules.

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Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

5 0
3 years ago
what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

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3 years ago
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Dimensional Analysis : 3 days to seconds
LuckyWell [14K]
Hi,

To convert 3 days to seconds write this.

1h = 3600s
24h = 3600 · 24 = 86400
3 days = 3 · 86400 = 259200sec

Hope this helps.
r3t40
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3 years ago
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1.) The object's Velocity

Faster it goes, more kinetic energy it has

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gladu [14]

v = u + at

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