Acceleration is the rate of change in an object's velocity
Answer: X = 52,314.12 N
Explanation: Let X be the force the feet of the athlete exerts on the floor.
According to newton's third law of motion the floor gives an upward reaction based on the weight of the athlete and the barbell which is known as the normal reaction ( based on the mass of the athlete and the barbell)
Mass of athlete = 87kg, mass of barbell = 600/ hence total normal reaction from the floor = 87* 61.22/ 9.8 *9.8 = 52,200N.
The athlete lifts the barbell from rest thus making it initial velocity u=0, distance covered = S = 0.65m and the time taken = 1.3s
The acceleration of the barbell is gotten by using the equation of constant acceleration motion
S= ut + 1/2at²
But u = 0
S = 1/2at²
0.65 = 1/2 *a (1.3)²
0.65 = 1.69 * a/2
0.65 * 2 = 1.69 * a
a = 0.65 * 2/ 1.69
a = 0.77m/s²
According to newton's second law of motion
Resultant force = mass * acceleration
And resultant force in this case is
X - 52,200 = (87 + 61.22) * 0.77
X - 52,200 = 148.22 * 0.77
X - 52, 200 = 114.132
X = 114.132 + 52,200
X = 52,314.12 N
T = 3.5 secs
Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
Answer:
5.571 sec
Explanation:
angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s
Period To = 2π / angular frequency
Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got
T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec
