The coil springs on a car's suspension have a value of k = 64000 N/m. When the

Physics

1 answer:

AlladinOne [14]3 years ago

8 0

80 joule is momentarily stored in each spring

Solution:

Given that,

The coil springs on a car's suspension have a value of k = 64000 N/m

When the car strikes a bump the springs briefly compress by 5.0 cm (.05 m)

By compressing the spring, we apply a force over a distance

As a result we have done work on the spring

Doing work means that we have transferred energy to spring in form of elastic potential

Therefore,

k = 64000 N/m

x = 0.05m

The elastic potential energy is given as:

$PE = \frac{1}{2}kx^2$

Where, "k" is the spring constant and "x" is the displacement

$PE = \frac{1}{2} \times 64000 \times 0.05^2\\\\PE = 32000 \times 0.0025\\\\PE = 80$

Thus 80 joule is momentarily stored in each spring

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A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

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3 years ago

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

$F_s = \mu_s*N$

Where $F_s$ is the static friction force, $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, $N = 110\text{ N}$ , to make the puppy moving we need to use a force of 80 N, therefore, $F_s = 80 \text{ N}$ , so we can solve for the coefficient as shown below: