The speed of an object can be determined from the distance vs time graph.
You know that speed = distance/time
in the graph, distance/time = slope of the curve.
So SPEED IS GIVEN BY THE SLOPE of the curve in the graph.
● If the distance vs time curve is a straight line, parallel to time axis(x-axis), slope is 0. That means speed is 0. So the object is at rest.
● If the distance vs time curve is a straight line, with some non-zero slope; That means speed is nonzero and constant. So the object is in uniform motion.
● If the distance vs time curve is a curved, the slope is changing. That means speed is changing. So the object is in an accelerated motion.
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
268N
Explanation:
The upward force acting on the block are the reaction and the hooked table..
The total normal force acting = normal reaction + 24N
Note that the normal reaction is always equal the weight of the table
Hence the normal force acting in the block is 244.0+24 = 268.0N
Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s²
to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg
