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3 years ago

8

An object of mass 9.00 kg attached to an ideal massless spring is pulled with a steady horizontal force across a frictionless le

vel surface. If the spring constant is 95.0 N/m and the spring is stretched by 10.0 cm , what is the magnitude of the acceleration of the object?

1 answer:

frosja888 [35]3 years ago

5 0

Answer:

Acceleration of the object will be $1.05m/sec^2$

Explanation:

We have given mass of the object m = 9 kg

spring constant k = 95 N/m

Spring is stretched by 10 cm

So x = 10 cm = 0.1 m

We know that force is given by $F=kx=95\times 0.1=9.5N$

From newton's law we also know that force is given by

F = ma , here m is mass and a is acceleration

So $9.5=9\times a$

$a=1.05m/sec^2$

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A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it

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2 years ago

Read 2 more answers

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

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4 years ago