5.22 - Interactive problem: forces on a sliding block
Above, you see an illustration of a block that is being pulled up a ramp by a rope. In
the simulation on the right, the force vectors on the block are drawn, but they are in
the wrong directions, have the wrong magnitudes, or both. Your job is to fix the
force vectors. If you do this correctly, the block will accelerate up the ramp at a rate
of 4.3 m/s^2. If not, the block will move due to the net force as determined by your
vectors as well as its mass.
The mass of the block is 6.0 kg. The amount of tension from the rope is 78 N and
the coefficient of kinetic friction is 0.45. The angle the ramp makes with the
horizontal is 30°. Calculate (to the nearest newton) the directions and magnitudes
of the weight, normal force and friction force. Drag the head of a vector to set its
magnitude and direction. You can also set the magnitudes in the control panel. The
vectors will “snap” to angles.
When you have arranged all the vectors, press the GO button. If your free-body
diagram is accurate, the block will accelerate up the ramp at 4.3 m/s^2. Press RESET
to try again.
There is more than one way to set the vectors to produce the same acceleration, but only one arrangement agrees with all the information
given. If you have difficulty solving this problem, review the sections on kinetic friction and the normal force, and the sample problem involving
a force at an angle.
A rope pulls the block up the ramp.
Draw a free-body diagram of the
forces on the block. If the diagram is
correct, the block will accelerate up
the ramp at 4.3 m/s^2.
5.23 - Hooke’s law and spring force
You probably already know a few basic things about springs: You stretch them, they
pull back on you. You compress them, they push back.
As a physics student, though, you are asked to study springs in a more quantitative
way. Let’s consider the force of a spring using the configuration shown in Concept 1.
Initially, no force is applied to the spring, so it is neither stretched nor compressed.
When no force is applied, the end of the spring is at a position called the rest point
(sometimes called the equilibrium point).
Then we stretch the spring. In the illustration to the right, the hand pulls to the right, so
the end of the spring moves to the right, away from its rest point, and the spring pulls
back to the left.
Hooke’s law is used to determine how much force the spring exerts. It states that the
amount of force is proportional to how far the end of the spring is stretched or
compressed away from its rest point. Stretch the end of the spring twice as far from its
rest point, and the amount of force is doubled.
The amount of force is also proportional to a spring constant, which depends on the
construction of the spring. A “stiff” spring has a greater spring constant than one that is
easier to stretch. Stiffer springs can be made from heavier gauge materials. The units
for spring constants are newtons per meter (N/m).
The equation for Hooke’s law is shown in Equation 1. The spring constant is represented by k. The displacement of the end of the spring is
represented by x. At the rest position, x = 0. When the spring is stretched, the displacement of the end of the spring has a positive x value.
When it is compressed, x is negative.
Hooke's law calculates the magnitude of the spring force. The equation has a negative sign to indicate that the force of a spring is a restoring
force, which means it acts to restore the end of the spring to its rest point. Stretch a spring and it will pull back toward the rest position;
compress a spring, and it will push back toward the rest position. The direction of the force is the opposite of the direction of the displacement.
Spring force
Force exerted by spring depends on:
·How much it is stretched or
compressed
·Spring constant