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
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
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