B. Transverse Wave this is the correct answer
Hi there!
To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.
Recall:
The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.
We can plug in the known values to solve for one part of the normal force:
N = (1)(9.8)(cos30) + F(.5) = 8.49 + .5F
Now, we can plug this into the equation for the dynamic friction force:
Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F
For a block to move with constant speed, the summation of forces must be equivalent to 0 N.
If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:
Fcosθ = 1.697 + .1F
Solve for F:
Fcos(30) - .1F = 1.697
F(cos(30) - .1) = 1.697
F = 2.216 N
Answer:
B
Explanation:
The kinetic energy in an object is converted into potential energy. This makes the kinetic decrease, while the potential increases.
The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency,
= 3.95 Hz
b). The period,
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
Hence, t = 0.0236 seconds.