# Work-Energy principle and spring force

• Nayef
In summary: The maximum height is the distance you pulled the weight down.In summary, a block on a frictionless surface with a spring constant of 50 N/m is initially at rest at position x = 0. When a constant force of 3 N is applied, the block moves in the positive direction of the x-axis until it reaches a stopping point. The position of the block at this point is x = 0.06 m. To find the position at which the block has maximum kinetic energy, we use the equation K = Fx - 1/2kx^2 and take the derivative of K with respect to x, setting it equal to zero. This gives us the position xc = 0.060 m at which
Nayef
A block lies on a horizontal frictionless surface, and the spring constant is 50 N/m. Initially, the spring
is at its relaxed length and the block is station ary at position x = 0 .Then an applied force with a constant magnitude of 3 N pulls the block in the positive direction of the x axis,stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) block's position when its kinetic energy is maximum

Ws(work done by spring force) = -Wapp(work done by applied force) , Fs(spring force) = -kx , Ws = -1/2 kx2I tried to solve the solution by using the concept , that because our applied force is of constant magnitude and because the block stops after some displacement x , then at that point Fapp = -Fs
==> Fapp = -(-kx) ==> 3= 50x ==> x = 3/50 = 0.06
But the answer and method according to the solution manual is different, it is given below
This is a situation where Eq Ws = -Wapp , applies, so we have

Fx = 1/2 kx2
==> 3 x = 1/2 (50 )x2
which (other than the zero) gives x = (3.0/25) m = 0.12 m.
What is wrong with my logic, the second part of the question is clear to me but i am providing its answer according to the solution manual because it seems relevant part first and might help in understanding part first better.

With Kf = K considered variable and Ki = 0, Eq. 7-27 gives K = Fx – 1/2kx2. We take
the derivative of K with respect to x and set the resulting expression equal to zero, in
order to find the position xc which corresponds to a maximum value of K:
xc = F/k= (3.0/50) m = 0.060 m.

Welcome to PF!
Nayef said:
A block lies on a horizontal frictionless surface, and the spring constant is 50 N/m. Initially, the spring
is at its relaxed length and the block is station ary at position x = 0 .Then an applied force with a constant magnitude of 3 N pulls the block in the positive direction of the x axis,stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) block's position when its kinetic energy is maximum

Ws(work done by spring force) = -Wapp(work done by applied force) , Fs(spring force) = -kx , Ws = -1/2 kx2I tried to solve the solution by using the concept , that because our applied force is of constant magnitude and because the block stops after some displacement x , then at that point Fapp = -Fs
==> Fapp = -(-kx) ==> 3= 50x ==> x = 3/50 = 0.06

But the answer and method according to the solution manual is different, it is given below
This is a situation where Eq Ws = -Wapp , applies, so we have

==> 3 x = 1/2 (50 )x2
which (other than the zero) gives x = (3.0/25) m = 0.12 m.
What is wrong with my logic, the second part of the question is clear to me but i am providing its answer according to the solution manual because it seems relevant part first and might help in understanding part first better.

With Kf = K considered variable and Ki = 0, Eq. 7-27 gives K = Fx – 1/2kx2. We take
the derivative of K with respect to x and set the resulting expression equal to zero, in
order to find the position xc which corresponds to a maximum value of K:
xc = F/k= (3.0/50) m = 0.060 m.

The block stops (momentarily) means that its velocity is zero at that moment. But it does not mean that the net force is zero; the acceleration is zero instead.
When only a constant force acts on the block, connected to the spring, it will perform SHM when no dissipative force is present. When it has maximum kinetic energy, its acceleration is zero, and when the kinetic energy is zero, the magnitude of acceracion is maximum.
You can do a little experiment: gravity also means a constant force. Hang some weight on an elastic spring. Holding the other end of the spring, let the weight fall : the weight will vibrate between a maximum and minimum height.

Last edited:
Nayef

## What is the Work-Energy principle?

The Work-Energy principle states that the work done by the sum of all forces acting on an object is equal to the change in the object's kinetic energy. In other words, if there is no net work done on an object, its kinetic energy will remain constant.

## What is the formula for calculating work?

The formula for calculating work is W = F * d * cos(theta), where W is work, F is the applied force, d is the displacement of the object, and theta is the angle between the force and the displacement vectors.

## How does the Work-Energy principle apply to spring force?

When a spring is compressed or stretched, it exerts a force on an object that is proportional to the displacement of the object from its equilibrium position. This force can be calculated using the formula F = -kx, where k is the spring constant and x is the displacement. The work done by the spring force is equal to the change in the object's potential energy, according to the Work-Energy principle.

## Can the Work-Energy principle be applied to non-conservative forces?

Yes, the Work-Energy principle can be applied to non-conservative forces, but it requires taking into account any non-conservative work done on the object. This includes work done by friction or air resistance, which can cause a decrease in the object's mechanical energy.

## What are the units for work and energy?

The SI unit for work and energy is the joule (J). However, other units such as foot-pounds (ft-lb) and calorie (cal) are also commonly used. The conversion between these units is: 1 J = 0.738 ft-lb = 0.239 cal.

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