A book is moved once around the perimeter of a tabletop with dimensions 1.2 m x 1.8 m.

1 answer:

3 0

Answer:

a) 0m

b) 6m

Explanation:

First, we need to remember:

Displacement: Difference between final and initial position.

Distance traveled: Total distance traveled.

a) If the final position is the same as the initial position, then:

final position = initial position

And we know that:

displacement = final position - initial position = 0

Then the displacement of the book is zero.

We can assume that the book traveled along the perimeter of the table.

The table is a rectangle of width 1.2m and length 1.8m

Remember that for a rectangle of width W and length L, the perimeter is:

P = 2*L + 2*W

Then the perimeter of the table is:

P = 2*1.2m + 2*1.8m = 6m

This means that the distance traveled by the book is 6 meters.

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Margaret [11]

The answer is number 1

7 0

3 years ago

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$