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Chapter 6: Problem 46

A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1600 . \mathrm{N} / \mathrm{m} .\) The clay compresses the spring a certain maximum amount before momentarily stopping a) Find the maximum compression of the spring. b) Find the total work done on the clay during the spring's compression.

Short Answer

Expert verified
Answer: To find the maximum compression of the spring (x), use the formula: $$ x = \sqrt{\frac{2(PE_i + KE_i)}{k}} $$ where \(PE_i\) is the initial potential energy of the clay, calculated as \(mgh\), and \(KE_i\) is the initial kinetic energy, calculated as \(\frac{1}{2}mv_i^2\). Then, to find the total work done on the clay during the spring's compression (W_total), use the formula: $$ W_{total} = PE_s - (PE_i + KE_i) $$ where \(PE_s\) is the elastic potential energy stored in the spring, calculated as \(\frac{1}{2}kx^2\). The negative sign indicates that the work is done against the motion of the clay to stop it.

Step by step solution

01

Calculate the initial potential energy and kinetic energy of the clay

Before the clay falls, it has potential energy due to its height and kinetic energy due to its initial speed. We can find both of these using the following formulas: $$ PE_i = mgh $$ $$ KE_i = \frac{1}{2}mv_i^2 $$ where \(PE_i\) is the initial potential energy, \(m\) is the mass of the clay, \(g\) is the acceleration due to gravity (approximately \(9.81ms^{-2}\)), \(h\) is the initial height, \(KE_i\) is the initial kinetic energy, and \(v_i\) is the initial speed of the clay.
02

Write the energy balance equation

When the clay compresses the spring to its maximum compression, it has transferred all its initial potential and kinetic energy into the elastic potential energy stored in the spring. So, we write the energy balance equation as follows: $$ PE_i + KE_i = PE_s $$ where \(PE_s\) is the elastic potential energy stored in the spring.
03

Calculate the maximum compression of the spring

We can use Hooke's law to calculate the elastic potential energy stored in the spring: $$ PE_s = \frac{1}{2}kx^2 $$ where k is the spring constant and x is the maximum compression of the spring. Substituting this into the energy balance equation from Step 2, we get: $$ PE_i + KE_i = \frac{1}{2}kx^2 $$ Solving for x, we get: $$ x = \sqrt{\frac{2(PE_i + KE_i)}{k}} $$ Plug in the values for \(PE_i\), \(KE_i\), and k from Steps 1 and 3, and calculate x.
04

Calculate the total work done on the clay during spring's compression

Because the conservation of mechanical energy holds during the process, the total work done on the clay during the spring's compression can be expressed as the change in mechanical energy: $$ W_{total} = PE_s - (PE_i + KE_i) $$ The negative sign indicates that the work is done against the motion of the clay to stop it. Calculate \(W_{total}\) using the values from Steps 1 and 3, and the formula for \(PE_s\) from Step 3.

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Most popular questions from this chapter

In 1896 in Waco, Texas, William George Crush, owner of the K-T (or "Katy") Railroad, parked two locomotives at opposite ends of a 6.4 -km-long track, fired them up, tied their throttles open, and then allowed them to crash head- on at full speed in front of 30,000 spectators. Hundreds of people were hurt by flying debris; several were killed. Assuming that each locomotive weighed \(1.2 \cdot 10^{6} \mathrm{~N}\) and its acceleration along the track was a constant \(0.26 \mathrm{~m} / \mathrm{s}^{2},\) what was the total kinetic energy of the two locomotives just before the collision?

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A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

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A 1.00 -kg block initially at rest at the top of a 4.00 -m incline with a slope of \(45.0^{\circ}\) begins to slide down the incline. The upper half of the incline is frictionless, while the lower half is rough, with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.300\). a) How fast is the block moving midway along the incline, before entering the rough section? b) How fast is the block moving at the bottom of the incline?
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