What is the acceleration of a 20-kg block of cement that is pulled upward with a force of 300 N?

Question

(300-N is not net force)

Answer 1

ID the forces: weight (W) pulling down, force = F pulling up.

Sum the forces: f = (W - F) = ma

Find a: a = (W - F)/m = (mg - F)/m = (20 X 9.81 - 300)/20 ~ -100/20 = -5 m/sec^2; where g = 9.81 m/sec^2 and a < 0 signifies the acceleration is upward because we assumed g > 0 was downward.

I used g ~ 10 m/sec^2 for ease of math, if you need more precision do the math using g = 9.81 m/sec^2.

Lesson: In all force-acceleration problems, first ID all the forces at work, then sum them. If the net force (f) <> 0, then the body all those forces are working on will be accelerating/decelerating. If f = 0, the body will be moving at a constant velocity, including v = 0.

Answer 2

Gravity pulls products downwards with a rigidity of 9.eighty one N/Kg. 50 Kg*9.eighty one N/Kg=490.5 N no remember if that's being pulled upwards with 600N, and downwards with 490.5 N, simply by fact of this common, it leads to an upwards rigidity of 109.5 N. As rigidity=Mass*Acceleration, 109.5 N=50 Kg *Acceleration 109.5/50 is the acceleration, it is approximately 2.2 metres in keeping with 2d in keeping with 2d.

Answer 3

F = 300 N
m = 20 kg
g = 9.81 m/s²

F = m*(a-g)

a - g = 300/20 = 15 m/s²

a = 15 - 9.81 = 5.19 m/s²

Answer 4

The second law of Newton states

F=mass *acceleration so acceleration=F/mass = 300/20=15m/s^2