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A block of mass =5.33 kg

is attached to a spring, which is resting on a horizontal frictionless table. The block is pushed into the spring, compressing it by 5.00 cm,
and is then released from rest. The spring begins to push the block back toward the equilibrium position at =0 cm.
The graph shows the component of the force (in Newtons) exerted by the spring on the block versus the position of the block (in centimeters) relative to equilibrium. Use the graph to answer the questions.
A force versus position graph, with F subscript X on the vertical axis in units of newtons, and X on the horizontal axis in units of centimeters. The tick marks on the F sub X axis appear every 1 newton, with 0 at the origin and ranging negative 6 newtons to 6 newtons. The tick marks on the X axis appear every 1 centimeter, with at 0 at the origin and ranging from negative 5 centimeters to 5 centimeters.
A straight, decreasing line segment on this graph has one endpoint at a X equals negative 5 centimeters and F sub X equals 6 newtons, and the other endpoint at X equals 5 centimeters and F sub X equals negative 6 newtons.

1) How much work
does the spring do pushing the block from its initial position at =−5.00 cm
to =2.75 cm?

2) What is the speed f
of the block when it reaches =2.75 cm?

3) What is the maximum speed max
of the block?

A block of mass =5.33 kg is attached to a spring, which is resting on a horizontal-example-1
User Cyberdenz
by
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1 Answer

5 votes

Answer:

Explanation:The work done by the spring pushing the block from its initial position at = −5.00 cm to = 2.75 cm is equal to the area under the force versus position curve between those two points. Since the graph is a straight line, we can use the formula for the area of a trapezoid:

Area = (1/2) * (b1 + b2) * h

where b1 and b2 are the lengths of the parallel sides of the trapezoid (i.e., the magnitudes of the forces at the initial and final positions), and h is the height of the trapezoid (i.e., the distance between the initial and final positions).

In this case, the initial position is = −5.00 cm and the final position is = 2.75 cm. From the graph, we can see that the magnitude of the force at the initial position is approximately 6 N, and the magnitude of the force at the final position is approximately 2 N (since the graph is symmetric, we could also use the negative values and obtain the same result).

So, the area of the trapezoid is:

Area = (1/2) * (6 N + 2 N) * (2.75 cm - (-5.00 cm)) = 21 N·cm

Therefore, the spring does 21 J of work pushing the block from = −5.00 cm to = 2.75 cm.

We can use conservation of energy to find the speed of the block when it reaches = 2.75 cm. At the initial position, the block is at rest and has zero kinetic energy. As the spring pushes the block back toward equilibrium, the potential energy stored in the spring is converted to kinetic energy of the block. At the final position, all of the potential energy has been converted to kinetic energy.

The potential energy stored in the spring at position x is given by:

U = (1/2) * k * (x - x0)^2

where k is the spring constant and x0 is the equilibrium position. In this case, x0 = 0 cm and k is not given, so we cannot calculate the potential energy directly. However, we know that the force exerted by the spring is proportional to the displacement from the equilibrium position, so we can write:

F = -kx

where the negative sign indicates that the force is in the opposite direction of the displacement (i.e., it is a restoring force). At the initial position, x = -5.00 cm and F = 6 N. At the final position, x = 2.75 cm and F = -2 N.

We can use the work-energy theorem to relate the work done by the spring to the change in kinetic energy of the block:

W = ΔK

where W is the work done by the spring and ΔK is the change in kinetic energy. Since the work done by the spring is the negative of the area under the force versus position curve between the initial and final positions (i.e., the same area we calculated in part 1), we know that W = -21 J. Therefore:

ΔK = -W = 21 J

At the final position, the block has kinetic energy equal to ΔK. We can use the formula for kinetic energy to find the speed v:

K = (1/2) * m * v^2

where m is the mass of the block. The mass is given as 5.33 kg, so we have:

21 J = (1/2) * 5.33 kg * v^2

Solving for v, we get:

v = sqrt((2

User Yet Another Geek
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